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+\section{Speed of Body under Free Fall from Height}
+Tags: Mechanics, Gravity, Speed of Body under Free Fall from Height
+
+\begin{theorem}
+Let an [[Definition:Object|object]] $B$ be released above ground from a point near the [[Definition:Earth|Earth's]] surface and allowed to fall freely.
+Let $B$ fall a [[Definition:Displacement|distance]] $s$.
+Then:
+:$v = \sqrt {2 g s}$
+where:
+:$v$ is the [[Definition:Speed|speed]] of $B$ after having fallen a [[Definition:Displacement|distance]] $s$
+:$g$ is the [[Acceleration Due to Gravity]] at the height through which $B$ falls.
+It is supposed that the distance $s$ is small enough that $g$ can be considered [[Definition:Constant|constant]] throughout.
+\end{theorem}
+
+\begin{proof}
+From [[Body under Constant Acceleration/Velocity after Distance|Body under Constant Acceleration: Velocity after Distance]]:
+:$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf g \cdot \mathbf s$
+All [[Definition:Dot Product|dot products]] are between pairs of [[Definition:Parallel Lines|parallel]] [[Definition:Vector|vectors]].
+Thus by [[Cosine Formula for Dot Product]]:
+:$v^2 = u^2 + 2 g s$
+Here the body falls from [[Definition:Stationary|rest]], so:
+:$\mathbf u = \mathbf 0$
+Thus:
+:$v^2 = 2 g s$
+and so:
+:$v = \sqrt {2 g s}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Acceleration is Second Derivative of Displacement with respect to Time]]:
+:$\mathbf g = \dfrac {\d^2 \mathbf s} {\d t^2}$
+[[Definition:Integration|Integrating]] with respect to $t$, and by definition of [[Definition:Velocity|velocity]]:
+:$\mathbf v = \dfrac {\d \mathbf s} {\d t} = \mathbf g t + \mathbf c_1$
+When $t = 0$, we have that $\mathrm c_1$ is the initial [[Definition:Velocity|velocity]] $\mathbf v_0$, and so:
+:$\mathbf v = \dfrac {\d \mathbf s} {\d t} = \mathbf g t + \mathbf v_0$
+[[Definition:Integration|Integrating]] with respect to $t$ again:
+:$\mathbf s = \dfrac {\mathbf g t^2} 2 + \mathbf v_0 t + \mathbf c_2$
+When $t = 0$, we have that $\mathrm c_2$ is the initial [[Definition:Displacement|displacement]] $\mathbf s_0$, and so:
+:$\mathbf s = \dfrac {\mathbf g t^2} 2 + \mathbf v_0 t + \mathbf s_0$
+We have that $B$ is released [[Definition:Stationary|at rest]] starting at $\mathbf s = \mathbf 0$.
+Thus $\mathbf v_0 = \mathbf s_0 = \mathbf 0$ and so:
+:$\mathbf s = \dfrac {\mathbf g t^2} 2$
+:$\mathbf v = \mathbf g t$
+By eliminating $t$ and taking the scalar quantities:
+:$v = \sqrt {2 g s}$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the [[Principle of Conservation of Energy]]:
+:$K + P = C$
+where:
+:$K$ is the [[Definition:Kinetic Energy|kinetic energy]] of $B$
+:$P$ is the [[Definition:Potential Energy|potential energy]] of $B$
+:$C$ is a [[Definition:Constant|constant]].
+Let the [[Definition:Mass|mass]] of $B$ be $m$.
+From [[Kinetic Energy of Motion]]:
+:$K = \dfrac {m v^2} 2$
+where $v$ is the [[Definition:Speed|speed]] of $B$.
+From [[Potential Energy of Position]]:
+:$P = m g s$
+where $s$ is the [[Definition:Distance|distance]] fallen by $B$.
+Since $B$ falls from [[Definition:Stationary|rest]], its initial [[Definition:Kinetic Energy|kinetic energy]] is zero.
+Having fallen a distance $s$, $B$ has lost [[Definition:Potential Energy|potential energy]] $m g s$.
+Therefore:
+:$\dfrac {m v^2} 2 = m g s$
+from which the result follows by dividing both sides by $m$ and extracting the [[Definition:Square Root|square root]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Terminal Velocity of Body under Fall Retarded Proportional to Velocity}
+Tags: Mechanics
+
+\begin{theorem}
+Let $B$ be a [[Definition:Body|body]] falling in a [[Definition:Gravitational Field|gravitational field]].
+Let $B$ be falling through a medium which exerts a resisting [[Definition:Force|force]] $k \mathbf v$ upon $B$ which is [[Definition:Proportion|proportional]] to the [[Definition:Velocity|velocity]] of $B$ relative to the medium.
+Then the [[Definition:Terminal Velocity|terminal velocity]] of $B$ is given by:
+:$v = \dfrac {g m} k$
+\end{theorem}
+
+\begin{proof}
+Let $B$ start from [[Definition:Stationary|rest]].
+From [[Motion of Body Falling through Air]], the [[Definition:Differential Equation|differential equation]] governing the motion of $B$ is given by:
+:$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \dfrac {\d \mathbf s} {\d t}$
+Dividing through by $m$ and setting $c = \dfrac k m$ gives:
+:$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \dfrac {\d \mathbf s} {\d t}$
+By definition of [[Definition:Velocity|velocity]]:
+:$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v$
+and so:
+{{begin-eqn}}
+{{eqn | l = \int \dfrac {\d \mathbf v} {\mathbf g - c \mathbf v}
+      | r = \int \rd t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac 1 c \map \ln {\mathbf g - c \mathbf v}
+      | r = t + c_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf g - c \mathbf v
+      | r = \mathbf c_2 e^{-c t}
+      | c = 
+}}
+{{end-eqn}}
+When $t = 0$ we have that $\mathbf v = 0$ and so:
+:$\mathbf c_2 = \mathbf g$
+Hence by taking [[Definition:Magnitude|magnitudes]]:
+:$v = \dfrac g c \paren {1 - e^{-c t} }$
+Since $c > 0$ it follows that $v \to \dfrac g c$ as $t \to \infty$.
+Thus in the limit:
+:$v = \dfrac g c = \dfrac {g m} k$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Approximate Motion of Pendulum}
+Tags: Mechanics, Pendulums
+
+\begin{theorem}
+Consider a [[Definition:Pendulum|pendulum]] consisting of a [[Definition:Pendulum Bob|bob]] whose [[Definition:Mass|mass]] is $m$, at the end of a [[Definition:Rod|rod]] of negligible [[Definition:Mass|mass]] of [[Definition:Length (Linear Measure)|length]] $a$.
+Let the [[Definition:Pendulum Bob|bob]] be pulled to one side so that the [[Definition:Rod|rod]] is at a small [[Definition:Angle|angle]] $\alpha$ (less than about $10 \degrees$ or $15 \degrees$) from [[Definition:Vertical|the vertical]] and then released.
+Let $T$ be the [[Definition:Period of Pendulum|period]] of the [[Definition:Pendulum|pendulum]], that is, the [[Definition:Time Period|time]] through which the [[Definition:Pendulum Bob|bob]] takes to travel from one end of its path to the other, and back again.
+Then:
+:$T = 2 \pi \sqrt {\dfrac a g}$
+where $g$ is the [[Acceleration Due to Gravity]].
+\end{theorem}
+
+\begin{proof}
+At a time $t$, let:
+:the [[Definition:Rod|rod]] be at an [[Definition:Angle|angle]] $\theta$ to the [[Definition:Vertical|the vertical]]
+:the [[Definition:Pendulum Bob|bob]] be travelling at a [[Definition:Speed|speed]] $v$
+:the [[Definition:Displacement|displacement]] of the [[Definition:Pendulum Bob|bob]] from where it is when the [[Definition:Rod|rod]] is [[Definition:Vertical Line|vertical]], along its line of travel, be $s$.
+:[[File:MotionOfPendulum.png|350px]]
+From [[Motion of Pendulum]], the equation of motion of the [[Definition:Pendulum Bob|bob]] is given by:
+:$\dfrac {a^2} 2 \paren {\dfrac {\d \theta} {\d t} }^2 = g a \paren {a \cos \theta - a \cos \alpha}$
+[[Definition:Differentiation|Differentiating]] {{WRT|Differentiation}} $t$ and simplifying yields:
+:$a \dfrac {\d^2 \theta} {\d t^2} = -g \sin \theta$
+For small [[Definition:Angle|angles]], $\sin \theta \approx \theta$ and so:
+:$\dfrac {\d^2 \theta} {\d t^2} + \dfrac g a \theta = 0$
+This is an instance of a [[Definition:Linear Second Order Ordinary Differential Equation|$2$nd order linear ODE]].
+From [[Linear Second Order ODE/y'' + k^2 y = 0|Linear Second Order ODE: $y'' + k^2 y = 0$]], this has the [[Definition:General Solution of Differential Equation|solution]]:
+:$\theta = c_1 \sin \sqrt {\dfrac g a} t + c_1 \cos \sqrt {\dfrac g a} t$
+We have at $t = 0$ that $\theta = \alpha$ and $\dfrac {\d \theta} {\d t} = 0$.
+This yields:
+:$\theta = \alpha \cos \sqrt {\dfrac g a} t$
+and so the [[Definition:Period of Pendulum|period]] of this [[Definition:Pendulum|pendulum]] is:
+:$T = 2 \pi \sqrt {\dfrac a g}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Space is Separable iff Density not greater than Aleph Zero}
+Tags: Denseness, Separable Spaces
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Then:
+:$T$ is [[Definition:Separable Space|separable]] {{iff}} $d \left({T}\right) \leq \aleph_0$
+where
+:$d \left({T}\right)$ denotes the [[Definition:Density of Topological Space|density]] of $T$,
+:$\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]].
+\end{theorem}
+
+\begin{proof}
+:$T$ is [[Definition:Separable Space|separable]]
+{{iff}}
+:there exists a [[Definition:Countable Set|countable]] [[Definition:Subset|subset]] of $T$ which is [[Definition:Everywhere Dense|dense]] by definition of [[Definition:Separable Space|separable space]]
+{{iff}}
+:there exists a [[Definition:Subset|subset]] $A$ of $T$ such that $A$ is [[Definition:Everywhere Dense|dense]] and exists an [[Definition:Injection|injection]] $A \to \N$ by definition of [[Definition:Countable Set|countable set]]
+{{iff}}
+:there exists a [[Definition:Subset|subset]] $A$ of $T$ such that $A$ is [[Definition:Everywhere Dense|dense]] and $\left\vert{A}\right\vert \leq \left\vert{\N}\right\vert$ by [[Injection iff Cardinal Inequality]]
+{{iff}}
+:there exists a [[Definition:Subset|subset]] $A$ of $T$ such that $A$ is [[Definition:Everywhere Dense|dense]] and $\left\vert{A}\right\vert \leq \aleph_0$ by [[Aleph Zero equals Cardinality of Naturals]]
+{{iff}}
+:$d \left({T}\right) \leq \aleph_0$ by definition of [[Definition:Density of Topological Space|density]]
+where $\left\vert{A}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $A$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Union of Separated Sets equals Union of Boundaries}
+Tags: Separated Sets, Boundaries
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $T$.
+Let $A$ and $B$ are [[Definition:Separated Sets|separated]].
+Then:
+:$\partial \left({ A \cup B }\right) = \partial A \cup \partial B$
+where
+: $\partial A$ denotes the [[Definition:Boundary (Topology)|boundary]] of $A$,
+: $A \cup B$ denotes the [[Definition:Set Union|union]] of $A$ and $B$.
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Separated Sets|separated sets]]:
+:$(1): \quad A^- \cap B = A \cap B^- = \varnothing$
+By [[Separated Sets are Disjoint]]:
+:$A \cap B = \varnothing$
+{{begin-eqn}}
+{{eqn | l = \partial A \cup \partial B
+      | r = \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)
+      | c = [[Union of Boundaries]]
+}}
+{{eqn | r = \partial \left({A \cup B}\right) \cup \varnothing \cup \left({\partial A \cap \partial B}\right)
+      | c = [[Boundary of Empty Set is Empty]]
+}}
+{{eqn | r = \partial \left({A \cup B}\right) \cup \left({\partial A \cap \partial B}\right)
+      | c = [[Union with Empty Set]]
+}}
+{{end-eqn}}
+We will proof that
+:$\partial A \cap \partial B \subseteq \partial \left({A \cup B}\right)$
+Let $x \in \partial A \cap \partial B$.
+Then by definition of [[Definition:Set Intersection|intersection]]:
+:$x \in \partial A \land x \in \partial B$
+Hence by [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$x \in A^- \cap \left({\complement_T \left({A}\right)}\right)^- \land x \in B^- \cap \left({\complement_T \left({B}\right)}\right)^-$
+where
+:$A^-$ denotes the [[Definition:Closure (Topology)|closure]] of $A$,
+:$\complement_T \left({A}\right) = T \setminus A$ denotes the [[Definition:Relative Complement|relative complement]] of $A$ in $T$.
+Then by definition of [[Definition:Set Intersection|intersection]]:
+:$x \in A^- \land x \in B^-$
+Therefore by definition of [[Definition:Empty Set|empty set]], [[Definition:Set Intersection|intersection]], and $(1)$:
+:$x \notin B \land x \notin A$
+Then by definition of [[Definition:Set Union|union]]:
+:$x \notin A \cup B$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \in \complement_T \left({A \cup B}\right)$
+By definition of [[Definition:Closure (Topology)|closure]]:
+:$\complement_T \left({A \cup B}\right) \subseteq \left({\complement_T \left({A \cup B}\right)}\right)^-$
+Then by definition of [[Definition:Subset|subset]]:
+:$x \in \left({\complement_T \left({A \cup B}\right)}\right)^-$
+By [[Closure of Finite Union equals Union of Closures]]:
+:$A^- \cup B^- = \left({A \cup B}\right)^-$
+By definition of [[Definition:Set Union|union]]:
+:$x \in \left({A \cup B}\right)^-$
+Then by definition of [[Definition:Set Intersection|intersection]]:
+:$x \in \left({A \cup B}\right)^- \cap \left({\complement_T \left({A \cup B}\right)}\right)^-$
+Thus  by [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$x \in \partial \left({A \cup B}\right)$
+This ends the proof of inclusion.
+Thus by [[Union with Superset is Superset]] the result:
+:$\partial A \cup \partial B = \partial \left({A \cup B}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Empty Set is Empty}
+Tags: Boundaries, Boundary of Empty Set is Empty
+
+\begin{theorem}
+Let $T$ be a [[Definition:Topological Space|topological space]].
+Then: 
+:$\partial_T \O = \O$
+where $\partial_T \O$ denotes the [[Definition:Boundary (Topology)|boundary]] in topology $T$ of $\O$.
+\end{theorem}
+
+\begin{proof}
+By [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$\partial_T \O = \O^- \cap \relcomp T \O^-$
+where $\O^-$ denotes the [[Definition:Closure (Topology)|closure]] of $\O$.
+By [[Closure of Empty Set is Empty Set]]:
+:$\O^- = \O$
+Thus the result follows by [[Intersection with Empty Set]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From [[Open and Closed Sets in Topological Space]], $\O$ is [[Definition:Clopen Set|clopen]] in $T$.
+The result follows from [[Set is Clopen iff Boundary is Empty]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Terminal Velocity of Body under Fall Retarded Proportional to Square of Velocity}
+Tags: Mechanics
+
+\begin{theorem}
+Let $B$ be a [[Definition:Body|body]] falling in a [[Definition:Gravitational Field|gravitational field]].
+Let $B$ be falling through a medium which exerts a resisting [[Definition:Force|force]] of [[Definition:Magnitude|magnitude]] $k v^2$ upon $B$ which is [[Definition:Proportion|proportional]] to the [[Definition:Square (Algebra)|square]] of the [[Definition:Velocity|velocity]] of $B$ relative to the medium.
+Then the [[Definition:Terminal Velocity|terminal velocity]] of $B$ is given by:
+:$v =  \sqrt {\dfrac {g m} k}$
+\end{theorem}
+
+\begin{proof}
+Let $B$ start from [[Definition:Stationary|rest]].
+The [[Definition:Differential Equation|differential equation]] governing the motion of $B$ is given by:
+:$m \dfrac {\d^2 \mathbf s} {\d t^2} = m \mathbf g - k \paren {\dfrac {\d \mathbf s} {\d t} }^2$
+Dividing through by $m$ and setting $c = \dfrac k m$ gives:
+:$\dfrac {\d^2 \mathbf s} {\d t^2} = \mathbf g - c \paren {\dfrac {\d \mathbf s} {\d t} }^2$
+By definition of [[Definition:Velocity|velocity]]:
+:$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v^2$
+for some [[Definition:Constant|constant]] $c$.
+and so taking [[Definition:Magnitude|magnitudes]] of the vector quantities:
+{{begin-eqn}}
+{{eqn | l = \int \dfrac {\d v} {g - c v^2}
+      | r = \int \rd t
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac 1 {2 c} \sqrt {\frac c g} \ln \paren {\frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v} }
+      | r = t + c_1
+      | c = [[Primitive of Reciprocal of a squared minus x squared/Logarithm Form|Primitive of $\dfrac 1 {a^2 - x^2}$: Logarithm Form]]
+}}
+{{end-eqn}}
+We have $v = 0$ when $t = 0$ which leads to $\ln 1 = 0 + c_1$ and thus $c_1 = 0$.
+Hence:
+{{begin-eqn}}
+{{eqn | l = \ln \paren {\frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v} }
+      | r = 2 t \sqrt {c g}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\sqrt {\frac c g} + v} {\sqrt {\frac c g} - v}
+      | r = e^{2 t \sqrt {c g} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {\frac c g} + v
+      | r = \paren {\sqrt {\frac c g} - v} e^{2 t \sqrt {c g} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = v \paren {e^{2 t \sqrt {c g} } + 1}
+      | r = v \paren {e^{2 t \sqrt {c g} } - 1} \sqrt {\frac g c}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = \sqrt {\frac g c} \paren {\frac {e^{2 t \sqrt {c g} } - 1} {e^{2 t \sqrt {c g} } + 1} }
+      | c = 
+}}
+{{eqn | r = \sqrt {\frac g c} \paren {\frac {1 - e^{-2 t \sqrt {c g} } } {1 + e^{-2 t \sqrt {c g} } } }
+      | c = 
+}}
+{{end-eqn}}
+Since $c > 0$ it follows that $v \to \sqrt {\dfrac g c}$ as $t \to \infty$.
+Thus in the limit:
+:$v = \sqrt {\dfrac g c} =  \sqrt {\dfrac {g m} k}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Escape Velocity of Projectile fired Upwards}
+Tags: Mechanics
+
+\begin{theorem}
+Let $P$ be a [[Definition:Planet|planet]].
+Let $P$ have an [[Acceleration Due to Gravity]] at its surface of $g$.
+Let $P$ have a [[Definition:Radius of Sphere|radius]] of $R$.
+Then the [[Definition:Escape Velocity|escape velocity]] of $P$ is given by:
+:$V = \sqrt {2 g R}$
+\end{theorem}
+
+\begin{proof}
+Let a projectile $B$ of [[Definition:Mass|mass]] $m$ be fired [[Definition:Vertical Line|vertically]] upwards from the surface of $P$ at such a [[Definition:Speed|speed]] that it escapes the [[Definition:Gravitational Field|gravitational field]] of $P$ completely.
+$F$ be the [[Definition:Force|force]] exerted on the projectile by the [[Definition:Gravitational Field|gravitational field]] of $P$.
+Let $x$ be the distance of $B$ from the surface of $P$ at [[Definition:Time|time]] $t$.
+We have:
+:$F = -\dfrac k {x^2}$
+where $k$ can be calculated at the surface of $P$ as:
+{{begin-eqn}}
+{{eqn | l = F
+      | r = m g
+      | c = 
+}}
+{{eqn | r = \dfrac k{R^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k
+      | r = R^2 m g
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$F = -\dfrac {m g R^2} {x^2}$
+Let the $B$ be travelling away from $P$ with [[Definition:Speed|speed]] $v$ at [[Definition:Time|time]] $t$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = F
+      | r = -m \frac {\d v} {\d t}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\frac {m g R^2} {x^2}
+      | r = m v \frac {\d v} {\d x}
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$\displaystyle -\int \frac {R^2 g} {x^2} \rd x = \int v \rd v$
+This gives:
+:$\dfrac {R^2 g} x = \dfrac {v^2} 2 + C$
+Now when $x = R$, $v = v_0$, the launch velocity, which leads to:
+:$C = R g - \dfrac {v_0^2} 2$
+and so:
+:$\dfrac {v^2} 2 = \dfrac {R^2 g} x - R g + \dfrac {v_0^2} 2$
+For $\dfrac {v^2} 2 > 0$ we need:
+:$\dfrac {v_0^2} 2 > R g \paren {1 - \dfrac R x}$
+whatever $x$ may be.
+So when $x \to \infty$ we have:
+:$\dfrac {v_0^2} 2 > R g$
+Hence:
+:$v_0 > \sqrt {2 R g}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Boundaries}
+Tags: Boundaries
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $S$.
+Then:
+:$\partial A \cup \partial B = \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$
+where $\partial A$ denotes the [[Definition:Boundary (Topology)|boundary]] of $A$.
+\end{theorem}
+
+\begin{proof}
+First we will prove that
+:$\partial A \subseteq \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$
+Let $x \in \partial A$.
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$x \notin \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$
+Then by definition of [[Definition:Set Union|union]]:
+:$x \notin \partial \left({A \cup B}\right) \land x \notin \partial \left({A \cap B}\right) \land x \notin \left({\partial A \cap \partial B}\right)$
+By [[Characterization of Boundary by Open Sets]]:
+:$\exists Q \in \tau: x \in Q \land \left({\left({A \cup B}\right) \cap Q = \varnothing \lor \complement_S \left({A \cup B}\right) \cap Q = \varnothing}\right)$
+By [[Intersection Distributes over Union]]. [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union|Complement of Union]]:
+:$\left({A \cap Q}\right) \cup \left({B \cap Q}\right) = \varnothing \lor \complement_S \left({ A }\right) \cap \complement_S \left({B}\right) \cap Q = \varnothing$
+Because $x \notin \left({\partial A \cap \partial B}\right)$ therefore by definition of [[Definition:Set Intersection|intersection]]:
+:$x \notin \partial B$
+By [[Characterization of Boundary by Open Sets]]:
+:$\exists U \in \tau: x \in U \land \left({B \cap U = \varnothing \lor \complement_S \left({B}\right) \cap U = \varnothing}\right)$
+As $x \in \partial A$ by [[Characterization of Boundary by Open Sets]]:
+:$A \cap Q \ne \varnothing$
+Then by [[Union is Empty iff Sets are Empty]]:
+:$\left({A \cap Q}\right) \cup \left({B \cap Q}\right) \ne \varnothing$
+Hence
+:$\complement_S \left({ A }\right) \cap \complement_S \left({B}\right) \cap Q = \varnothing$
+We will show that
+:$B \cap U = \varnothing \implies \complement_S \left({ A }\right) \cap Q \cap U = \varnothing$
+Assume
+:$B \cap U = \varnothing$
+Then:
+:$U \subseteq \complement_S \left({B}\right)$
+By [[Intersection with Empty Set]]:
+:$\complement_S \left({ A }\right) \cap Q \cap \complement_S \left({B}\right) \cap U = \varnothing \cap U = \varnothing$
+Thus by [[Intersection with Subset is Subset]]:
+:$\complement_S \left({ A }\right) \cap Q \cap U = \varnothing$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$x \in Q \cap U$
+By definition of [[Definition:Topological Space|topological space]]:
+:$Q \cap U$ is [[Definition:Open Set (Topology)|open]].
+Then by [[Characterization of Boundary by Open Sets]]:
+:$\complement_S \left({A}\right) \cap Q \cap U \ne \varnothing$
+Hence:
+:$B \cap U \ne \varnothing$
+Then:
+:$\complement_S \left({B}\right) \cap U = \varnothing$
+Therefore:
+:$U \subseteq B$
+Because $x \notin \partial \left({A \cap B}\right)$ by [[Characterization of Boundary by Open Sets]]:
+:$\exists V \in \tau: x \in V \land \left({A \cap B \cap V = \varnothing \lor \complement_S \left({A \cap B}\right) \cap V = \varnothing}\right)$
+By [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|Complement of Intersection]]:
+:$A \cap B \cap V = \varnothing \lor \left({\complement_S \left({A}\right) \cup \complement_S \left({B}\right)}\right) \cap V = \varnothing$
+By [[Intersection Distributes over Union]]:
+:$A \cap V \cap B = \varnothing \lor \left({\complement_S \left({A}\right) \cap V}\right) \cup \left({\complement_S \left({B}\right) \cap V}\right) = \varnothing$
+Because $x \in \partial A$ therefore by [[Characterization of Boundary by Open Sets]]:
+:$\complement_S \left({A}\right) \cap V \ne \varnothing$
+Then by [[Union is Empty iff Sets are Empty]]:
+:$\left({\complement_S \left({A}\right) \cap V}\right) \cup \left({\complement_S \left({B}\right) \cap V}\right) \ne \varnothing$
+Hence:
+:$A \cap V \cap B = \varnothing$
+By [[Intersection with Empty Set]]:
+:$A \cap V \cap B \cap U = \varnothing \cap U = \varnothing$
+By [[Intersection with Subset is Subset]]:
+:$A \cap V \cap U = \varnothing$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$x \in V \cap U$
+By definition of [[Definition:Topological Space|topological space]]:
+:$V \cap U$ is [[Definition:Open Set (Topology)|open]].
+Then by [[Characterization of Boundary by Open Sets]]:
+:$A \cap Q \cap V \ne \varnothing$
+This contradicts with $A \cap V \cap U = \varnothing$
+Thus the inclusion is proved.
+{{qed|lemma}}
+Analogically 
+:$\partial B \subseteq \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$
+By [[Union of Subsets is Subset]]:
+:$\partial A \cup \partial B \subseteq \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$
+By [[Boundary of Union is Subset of Union of Boundaries]]:
+:$\partial \left({A \cup B}\right) \subseteq \partial A \cup \partial B$
+By [[Boundary of Intersection is Subset of Union of Boundaries]]:
+:$\partial \left({A \cap B}\right) \subseteq \partial A \cup \partial B$
+By [[Intersection is Subset of Union]]:
+:$\partial A \cap \partial B \subseteq \partial A \cup \partial B$
+Hence by [[Union of Subsets is Subset]]
+:$\partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right) \subseteq \partial A \cup \partial B$
+Thus by definition of [[Definition:Set Equality|set equality]] the result follows:
+:$\partial A \cup \partial B = \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Arch of Sine Function}
+Tags: Sine Function, Elliptic Integrals
+
+\begin{theorem}
+The [[Definition:Length of Curve|length]] of one [[Definition:Arch of Sine Function|arch]] of the [[Definition:Sine Function|sine function]]:
+:$y = \sin x$
+is given by:
+:$L = 2 \sqrt 2 \map E {\dfrac {\sqrt 2} 2}$
+where $E$ denotes the [[Definition:Complete Elliptic Integral of the Second Kind|incomplete elliptic integral of the second kind]].
+\end{theorem}
+
+\begin{proof}
+Let $L$ be the [[Definition:Length of Curve|length]] of one [[Definition:Arch of Sine Function|arch]] of $y = \sin x$.
+Then:
+{{begin-eqn}}
+{{eqn | l = L
+      | r = 2 \int_0^{\pi/2} \sqrt {1 + \paren {\map {\frac \d {\d x} } {\sin x} }^2} \rd x
+      | c = {{Defof|Length of Curve}}
+}}
+{{eqn | r = 2 \int_0^{\pi/2} \sqrt {1 + \cos^2 x} \rd x
+      | c = [[Derivative of Sine Function]]
+}}
+{{eqn | r = 2 \int_0^{\pi/2} \sqrt {2 - \sin^2 x} \rd x
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = 2 \sqrt 2 \int_0^{\pi/2} \sqrt {1 - \paren {\frac {\sqrt 2} 2} \sin^2 x} \rd x
+      | c = extracting $\sqrt 2$
+}}
+{{eqn | r = 2 \sqrt 2 \map E {\dfrac {\sqrt 2} 2}
+      | c = {{Defof|Complete Elliptic Integral of the Second Kind}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Lemniscate of Bernoulli}
+Tags: Lemniscates, Elliptic Integrals
+
+\begin{theorem}
+The total [[Definition:Length of Curve|length]] of the [[Definition:Lemniscate of Bernoulli|lemniscate of Bernoulli]] given in [[Definition:Polar Coordinates|polar coordinates]] as:
+:$r^2 = a^2 \cos 2 \theta$
+is given by:
+{{begin-eqn}}
+{{eqn | l = L
+      | r = 4 a \map F {\sqrt 2, \dfrac \pi 4}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {\sqrt {2 \pi} } \paren {\map \Gamma {\dfrac 1 4} }^2
+      | c = 
+}}
+{{end-eqn}}
+where $F$ denotes the [[Definition:Incomplete Elliptic Integral of the First Kind|incomplete elliptic integral of the first kind]].
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Length of Curve|arc length]] of a small length increment $\d s$ is given in [[Definition:Polar Coordinates|polar co-ordinates]] by:
+:$\paren {\d s}^2 = \paren {r \d \theta}^2 + \paren {\d r}^2$
+from which:
+:$\dfrac {\d s} {\d \theta} = \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2}$
+Half of one [[Definition:Lobe of Lemniscate of Bernoulli|lobe]] of the [[Definition:Lemniscate of Bernoulli|lemniscate]] is achieved when $\theta$ goes from $0$ to $\pi / 4$.
+Therefore the total [[Definition:Length of Curve|length]] of the [[Definition:Lemniscate of Bernoulli|lemniscate of Bernoulli]] is given by:
+:$\displaystyle L = 4 \int_0^{\pi/4} \sqrt {r^2 + \paren {\dfrac {\d r} {\d \theta} }^2} \rd \theta$
+First we show:
+{{begin-eqn}}
+{{eqn | l = r^2
+      | r = a^2 \cos 2 \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 r \frac {\d r} {\d \theta}
+      | r = -2 a^2 \sin 2 \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d r} {\d \theta}
+      | r = -\frac {a^2 \sin 2 \theta} {a \sqrt {\cos 2 \theta} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {\dfrac {\d r} {\d \theta} }^2
+      | r = \frac {a^2 \sin^2 2 \theta} {\cos 2 \theta}
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = r^2 + \paren {\dfrac {\d r} {\d \theta} }^2
+      | r = a^2 \cos 2 \theta + \frac {a^2 \sin^2 2 \theta} {\cos 2 \theta}
+      | c = 
+}}
+{{eqn | r = a^2 \frac {\cos^2 2 \theta + \sin^2 2 \theta} {\cos 2 \theta}
+      | c = 
+}}
+{{eqn | r = \frac {a^2} {\cos 2 \theta}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = L
+      | r = 4 a \int_0^{\pi / 4} \frac {\d \theta} {\sqrt {\cos 2 \theta} }
+      | c = 
+}}
+{{eqn | r = 4 a \int_0^{\pi / 4} \frac {\d \theta} {\sqrt {1 - 2 \sin^2 \theta} }
+      | c = 
+}}
+{{eqn | r = 4 a \map F {\sqrt 2, \dfrac \pi 4}
+      | c = {{Defof|Incomplete Elliptic Integral of the First Kind}}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Union is Subset of Union of Boundaries}
+Tags: Boundaries
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $S$.
+Then:
+:$\partial \left({A \cup B}\right) \subseteq \partial A \cup \partial B$
+where $\partial A$ denotes the [[Definition:Boundary (Topology)|boundary]] of $A$.
+\end{theorem}
+
+\begin{proof}
+By [[Intersection is Subset]]:
+:$\complement_S \left({A}\right) \cap \complement_S \left({B}\right) \subseteq \complement_S \left({A}\right) \land \complement_S \left({A}\right) \cap \complement_S \left({B}\right) \subseteq \complement_S \left({B}\right)$
+Then by [[Topological Closure of Subset is Subset of Topological Closure]]:
+:$\left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \subseteq \left({\complement_S \left({A}\right)}\right)^- \land \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \subseteq \left({\complement_S \left({B}\right)}\right)^-$
+Hence by [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$\left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap A^- \subseteq \partial A \land \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap B^- \subseteq \partial B$
+Thus
+{{begin-eqn}}
+{{eqn | l = \partial \left({A \cup B}\right)
+      | r = \left({\complement_S \left({A \cup B}\right)}\right)^- \cap \left({A \cup B}\right)^-
+      | c = [[Boundary is Intersection of Closure with Closure of Complement]]
+}}
+{{eqn | r = \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap \left({A \cup B}\right)^-
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union|Complement of Union]]
+}}
+{{eqn | r = \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap \left({A^- \cup B^-}\right)
+      | c = [[Closure of Finite Union equals Union of Closures]]
+}}
+{{eqn | r = \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap A^- \cup \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap B^-
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | o = \subseteq
+      | r = \partial A \cup \partial B
+      | c = [[Set Union Preserves Subsets]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Boundary of Intersection is Subset of Union of Boundaries}
+Tags: Boundaries
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $S$.
+Then:
+:$\partial \paren {A \cap B} \subseteq \partial A \cup \partial B$
+where $\partial A$ denotes the [[Definition:Boundary (Topology)|boundary]] of $A$.
+\end{theorem}
+
+\begin{proof}
+By [[Intersection is Subset]]:
+:$A \cap B \subseteq A \land A \cap B \subseteq B$
+Then by [[Topological Closure of Subset is Subset of Topological Closure]]:
+:$\paren {A \cap B}^- \subseteq A^- \land \paren {A \cap B}^- \subseteq B^-$
+Hence by [[Boundary is Intersection of Closure with Closure of Complement]]:
+:$\paren {A \cap B}^- \cap \paren {\relcomp S A}^- \subseteq \partial A \land \paren {A \cap B}^- \cap \paren {\relcomp S B}^- \subseteq \partial B$
+Thus
+{{begin-eqn}}
+{{eqn | l = \partial \paren {A \cap B}
+      | r = \paren {A \cap B}^- \cap \paren {\relcomp S {A \cap B} }^-
+      | c = [[Boundary is Intersection of Closure with Closure of Complement]]
+}}
+{{eqn | r = \paren {A \cap B}^- \cap \paren {\relcomp S A \cup \relcomp S B}^-
+      | c = [[De Morgan's Laws (Set Theory)/Relative Complement/Complement of Intersection|Complement of Intersection]]
+}}
+{{eqn | r = \paren {A \cap B}^- \cap \paren {\paren {\relcomp S A}^- \cup \paren {\relcomp S B}^-}
+      | c = [[Closure of Finite Union equals Union of Closures]]
+}}
+{{eqn | r = \paren {A \cap B}^- \cap \paren {\relcomp S A}^- \cup \paren {A \cap B}^- \cap \paren {\relcomp S B}^-
+      | c = [[Intersection Distributes over Union]]
+}}
+{{eqn | o = \subseteq
+      | r = \partial A \cup \partial B
+      | c = [[Set Union Preserves Subsets]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complete Elliptic Integral of the First Kind as Power Series}
+Tags: Elliptic Integrals
+
+\begin{theorem}
+The '''[[Definition:Complete Elliptic Integral of the First Kind|complete elliptic integral of the first kind]]''':
+:$\displaystyle \map K k = \int_0^{\pi / 2} \frac {\rd \phi} {\sqrt {1 - k^2 \sin^2 \phi} } = \int_0^1 \frac {\rd v} {\sqrt {\paren {1 - v^2} \paren {1 - k^2 v^2} } }$
+can be expressed as the [[Definition:Power Series|power series]]:
+{{begin-eqn}}
+{{eqn | l = \map K k
+      | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }^2 k^{2 i}
+      | c = 
+}}
+{{eqn | r = \frac \pi 2 \paren {1 + \paren {\frac 1 2}^2 k^2 + \paren {\frac {1 \cdot 3} {2 \cdot 4} }^2 k^4 + \paren {\frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} }^2 k^6 + \cdots}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Integral of Power of Sine]], $\forall i \in \N$:
+{{begin-eqn}}
+{{eqn | l = \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi
+      | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi - \intlimits {\frac {\sin^{2 i - 1} x \cos x} i} {x = 0} {x = \frac \pi 2}
+}}
+{{eqn | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi
+      | c = $\sin 0 = \cos \dfrac \pi 2 = 0$
+}}
+{{eqn | r = \cdots
+}}
+{{eqn | r = \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} \int_0^{\pi / 2} 1 \rd \phi
+}}
+{{eqn | r = \frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j}
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map K k
+      | r = \int_0^{\pi / 2} \frac {\rd \phi} {\sqrt{1 - k^2 \sin^2 \phi} }
+}}
+{{eqn | r = \int_0^{\pi / 2} \sum_{i \mathop \ge 0} \binom {- \frac 1 2} i \paren {- k^2 \sin^2 \phi}^i \rd \phi
+      | c = [[General Binomial Theorem]]
+}}
+{{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {\frac 1 2 - j} j} \paren {-1}^i \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi
+      | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}}
+}}
+{{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {1 - 2 j} {2 j} } \paren {\prod_{j \mathop = 1}^i \paren {-1} }\paren {\frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }
+      | c = from above
+}}
+{{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }^2 k^{2 i}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Complete Elliptic Integral of the Second Kind as Power Series}
+Tags: Elliptic Integrals
+
+\begin{theorem}
+The '''[[Definition:Complete Elliptic Integral of the Second Kind|complete elliptic integral of the second kind]]''':
+:$\displaystyle \map E k = \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \phi} \, \rd \phi = \int_0^1 \dfrac {\sqrt {1 - k^2 v^2} } {\sqrt {1 - v^2}} \, \rd v$
+can be expressed as the [[Definition:Power Series|power series]]:
+{{begin-eqn}}
+{{eqn | l = \map E k
+      | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 \frac {k^{2 i} } {1 - 2 i}
+      | c = 
+}}
+{{eqn | r = \frac \pi 2 \paren {1 - \paren {\frac 1 2}^2 k^2 - \paren {\frac {1 \cdot 3} {2 \cdot 4} }^2 \frac {k^4} 3 - \paren {\frac {1 \cdot 3 \cdot 5} {2 \cdot 4 \cdot 6} }^2 \frac {k^6} 5 - \cdots}
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+From [[Reduction Formula for Integral of Power of Sine]], $\forall i \in \N$:
+{{begin-eqn}}
+{{eqn | l = \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi
+      | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi - \intlimits {\frac {\sin^{2 i - 1} x \cos x} i} {x = 0} {x = \frac \pi 2}
+}}
+{{eqn | r = \frac {2 i - 1} {2 i} \int_0^{\pi / 2} \sin^{2 i - 2} \phi \rd \phi
+      | c = $\sin 0 = \cos \dfrac \pi 2 = 0$
+}}
+{{eqn | r = \cdots
+}}
+{{eqn | r = \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} \int_0^{\pi / 2} 1 \rd \phi
+}}
+{{eqn | r = \frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j}
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map E k
+      | r = \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \phi} \, \rd \phi
+}}
+{{eqn | r = \int_0^{\pi / 2} \sum_{i \mathop \ge 0} \binom {\frac 1 2} i \paren {- k^2 \sin^2 \phi}^i \rd \phi
+      | c = [[General Binomial Theorem]]
+}}
+{{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {\frac 3 2 - j} j} \paren {-1}^i \int_0^{\pi / 2} \sin^{2 i} \phi \rd \phi
+      | c = {{Defof|Binomial Coefficient|subdef = Real Numbers}}
+}}
+{{eqn | r = \sum_{i \mathop \ge 0} k^{2 i} \paren {\prod_{j \mathop = 1}^i \frac {3 - 2 j} {2 j} } \paren {\prod_{j \mathop = 1}^i \paren {-1} }\paren {\frac \pi 2 \prod_{j \mathop = 1}^i \frac {2 j - 1} {2 j} }
+      | c = From above
+}}
+{{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 3} {2 j} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} } k^{2 i}
+      | c = 
+}}
+{{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 3} {2 j} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} } k^{2 i}
+      | c = 
+}}
+{{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\frac {2 - 3} {2 i - 1} } \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 k^{2 i}
+      | c = 
+}}
+{{eqn | r = \frac \pi 2 + \frac \pi 2 \sum_{i \mathop \ge 1} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 \frac {k^{2 i} } {1 - 2 i}
+      | c = 
+}}
+{{eqn | r = \frac \pi 2 \sum_{i \mathop \ge 0} \paren {\prod_{j \mathop = 1}^i \frac {2 j - 1} { 2 j} }^2 \frac {k^{2 i} } {1 - 2 i}
+      | c = Product equals $1$ when $i = 0$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Discrete Space is Separable iff Countable}
+Tags: Denseness, Separable Spaces, Discrete Topology
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Discrete Space|discrete]] [[Definition:Topological Space|topological space]].
+Then:
+:$T$ is [[Definition:Separable Space|separable]] {{iff}} $S$ is [[Definition:Countable Set|countable]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Immediate from [[Separable Discrete Space is Countable]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Immediate from [[Countable Space is Separable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set is Countable iff Cardinality not greater Aleph Zero}
+Tags: Countable Sets
+
+\begin{theorem}
+Let $X$ be a [[Definition:Set|set]].
+Then:
+:$X$ is [[Definition:Countable Set|countable]] {{iff}} $\size X \le \aleph_0$
+where
+:$\size X$ denotes the [[Definition:Cardinality|cardinality]] of $X$,
+:$\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]].
+\end{theorem}
+
+\begin{proof}
+:$X$ is [[Definition:Countable Set|countable]]
+{{iff}}
+:there exists an [[Definition:Injection|injection]] $X \to \N$ by definition of [[Definition:Countable Set|countable set]]
+{{iff}}
+:$\size X \le \size \N$ by [[Injection iff Cardinal Inequality]]
+{{iff}}
+:$\size X \le \aleph_0$ by [[Aleph Zero equals Cardinality of Naturals]].
+{{qed}}
+[[Category:Countable Sets]]
+odwikgixvfndjekh3c1vm2d04dez9q6
+\end{proof}<|endoftext|>
+\section{Bernoulli's Theorem}
+Tags: Probability Theory
+
+\begin{theorem}
+Let the [[Definition:Probability|probability]] of the [[Definition:Occurrence of Event|occurrence]] of an [[Definition:Event|event]] be $p$.
+Let $n$ [[Definition:Independent Events|independent trials]] be made, with $k$ [[Definition:Success|successes]].
+Then:
+:$\displaystyle \lim_{n \mathop \to \infty} \frac k n = p$
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Random Variable|random variable]] $k$ have the [[Definition:Binomial Distribution|binomial distribution]] with parameters $n$ and $p$, that is:
+:$k \sim \Binomial n p$
+where $k$ denotes the number of [[Definition:Success|successes]] of the $n$ [[Definition:Independent Events|independent trials]] of the event with [[Definition:Probability|probability]] $p$.
+From [[Expectation of Binomial Distribution]]:
+:$\expect k = n p \leadsto \dfrac 1 n \expect k = p$
+[[Linearity of Expectation Function]] gives:
+:$ \expect {\dfrac k n} = p =: \mu$
+Similarly, from [[Variance of Binomial Distribution]]:
+:$\var k = n p \paren {1 - p} \leadsto \dfrac 1 {n^2} \var k = \dfrac {p \paren {1 - p} } n$
+From [[Variance of Linear Combination of Random Variables]]:
+:$\var {\dfrac k n} = \dfrac {p \paren {1 - p} } n =: \sigma^2$
+By applying [[Chebyshev's Inequality]] to $\dfrac {k} {n}$, we have for any $l>0$:
+:$\map \Pr {\size {\dfrac k m - \mu} \ge l \sigma} \le \dfrac 1 {l^2}$
+Now, let $\epsilon > 0$ and choose $l = \dfrac \epsilon \sigma$, to get:
+:$\map \Pr {\size {\dfrac k m - \mu} \ge \dfrac \epsilon \sigma \cdot \sigma} \le \dfrac {\sigma^2} {\epsilon^2}$
+Simplifying and plugging in the values of $\mu$ and $\sigma^2$ defined above yields:
+:$\map \Pr {\size {\dfrac k n - p} \ge \epsilon} \le \dfrac {p \paren {1 - p} } {n \epsilon^2}$
+Scaling both sides by $-1$ and adding $1$ to both sides yields:
+:$1 - \map \Pr {\size {\dfrac k n - p} \ge \epsilon} \ge 1 - \dfrac {p \paren {1 - p} } {n \epsilon^2}$
+Applying [[Union of Event with Complement is Certainty]] to the left hand side:
+:$\map \Pr {\size {\dfrac k n - p} \le \epsilon} \ge 1 - \dfrac {p \paren {1 - p} } {n\epsilon^2}$
+Taking the limit as $n$ approaches infinity on both sides, we have:
+:$\displaystyle \lim_{n \mathop \to \infty} \map \Pr {\size {\frac k n - p} < \epsilon} = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Time of Travel down Brachistochrone}
+Tags: Cycloids
+
+\begin{theorem}
+Let a [[Definition:Wire|wire]] $AB$ be curved into the shape of a [[Definition:Brachistochrone|brachistochrone]].
+Let $AB$ be embedded in a [[Definition:Constant|constant]] and [[Definition:Uniform|uniform]] [[Definition:Gravitational Field|gravitational field]] where [[Acceleration Due to Gravity]] is $g$.
+Let a [[Definition:Bead|bead]] $P$ be released at $A$ to slide down without [[Definition:Friction|friction]] to $B$.
+Then the [[Definition:Time|time]] taken for $P$ to slide from $A$ to $B$ is:
+:$T = \pi \sqrt {\dfrac a g}$
+where $a$ is the [[Definition:Radius of Circle|radius]] of the [[Definition:Generating Circle of Cycloid|generating circle]] of the [[Definition:Cycloid|cycloid]] which forms $AB$.
+\end{theorem}
+
+\begin{proof}
+That the [[Definition:Curve|curve]] $AB$ is indeed a [[Definition:Cycloid|cycloid]] is demonstrated in [[Brachistochrone is Cycloid]].
+Let $A$ be located at the [[Definition:Origin|origin]] of a [[Definition:Cartesian Plane|cartesian plane]].
+We have the equations of the [[Definition:Cycloid|cycloid]]:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = a \paren {\theta - \sin \theta}
+}}
+{{eqn | l = y
+      | r = a \paren {1 - \cos \theta}
+}}
+{{end-eqn}}
+[[Definition:Differentiation With Respect To|Differentiating with respect to]] theta:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d \theta}
+      | r = \paren {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | l = \frac {\d y} {\d \theta}
+      | r = a \sin \theta
+}}
+{{end-eqn}}
+Let $s$ be the [[Definition:Distance (Linear Measure)|distance]] along $AB$ from the [[Definition:Origin|origin]].
+Then:
+{{begin-eqn}}
+{{eqn | l = \paren {\frac {\d s} {\d \theta} }^2
+      | r = \paren {\frac {\d x} {\d \theta} }^2 + \paren {\frac {\d y} {\d \theta} }^2
+      | c = 
+}}
+{{eqn | r = a^2 \paren {1 - \cos \theta}^2 + a^2 \sin^2 \theta
+      | c = 
+}}
+{{eqn | r = a^2 \paren {1 - 2 \cos \theta + \cos^2 \theta + \sin^2 \theta}
+      | c = 
+}}
+{{eqn | r = 2 a^2 \paren {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | r = 4 a^2 \map {\sin^2} {\frac \theta 2}
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\dfrac {\d s} {\d \theta} = 2 a \map \sin {\dfrac \theta 2}$
+Now:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \dfrac {\d s} {\d t}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int \rd t
+      | r = t
+      | c = 
+}}
+{{eqn | r = \int \frac {\d s} v
+      | c = 
+}}
+{{end-eqn}}
+We also have, from the [[Principle of Conservation of Energy]], that:
+{{begin-eqn}}
+{{eqn | l = m g y
+      | r = \frac {m v^2} 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = v
+      | r = \dfrac {\d s} {\d t}
+      | c = 
+}}
+{{eqn | r = \sqrt {2 g y}
+      | c = 
+}}
+{{eqn | r = \sqrt {2 g a \paren {1 - \cos \theta} }
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \frac {\d t} {\d \theta}
+      | r = \frac {\d t} {\d s} \theta
+      | c = 
+}}
+{{eqn | r = \frac {a \sqrt {2 \paren {1 - \cos \theta} } } {\sqrt {2 g a \paren {1 - \cos \theta} } }
+      | c = 
+}}
+{{eqn | r = \sqrt {\frac a g}
+      | c = 
+}}
+{{end-eqn}}
+Let $T$ be the time taken to slide down the [[Definition:Brachistochrone|brachistochrone]].
+At the top:
+:$\theta = 0$
+and at the bottom:
+:$\theta = \pi$
+Then:
+{{begin-eqn}}
+{{eqn | l = T
+      | r = \int \rd t
+      | c = 
+}}
+{{eqn | r = \sqrt {\frac a g} \int_0^\pi \rd \theta
+      | c = 
+}}
+{{eqn | r = \pi \sqrt {\frac a g}
+      | c = 
+}}
+{{end-eqn}}
+So the [[Definition:Time|time]] to slide down this [[Definition:Brachistochrone|brachistochrone]] from the top to the bottom is:
+:$T = \pi \sqrt {\dfrac a g}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Time of Travel down Brachistochrone/Corollary}
+Tags: Cycloids
+
+\begin{theorem}
+Let a [[Definition:Bead|bead]] $P$ be released from anywhere on the [[Definition:Wire|wire]] between $A$ and $B$ to slide down without [[Definition:Friction|friction]] to $B$.
+Then the [[Definition:Time|time]] taken for $P$ to slide to $B$ is:
+:$T = \pi \sqrt{\dfrac a g}$
+\end{theorem}
+
+\begin{proof}
+That the [[Definition:Curve|curve]] $AB$ is indeed a [[Definition:Cycloid|cycloid]] is demonstrated in [[Brachistochrone is Cycloid]].
+Let $A$ be located at the [[Definition:Origin|origin]] of a [[Definition:Cartesian Plane|cartesian plane]].
+Let the [[Definition:Bead|bead]] slide from an intermediate point $\theta_0$.
+We have:
+:$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$
+which leads us, via the same route as for [[Time of Travel down Brachistochrone]], to:
+{{begin-eqn}}
+{{eqn | l = T
+      | r = \int_{\theta_0}^\pi \sqrt {\frac {2 a^2 \paren {1 - \cos \theta} } {2 g a \paren {\cos \theta_0 - \cos \theta} } } \rd \theta
+      | c = 
+}}
+{{eqn | r = \sqrt {\frac a g} \int_{\theta_0}^\pi \sqrt {\frac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta
+      | c = 
+}}
+{{end-eqn}}
+Using the [[Half Angle Formula for Cosine]] and [[Half Angle Formula for Sine]], this gives:
+:$\displaystyle T = \sqrt{\frac a g} \int_{\theta_0}^\pi \frac {\map \sin {\theta / 2} } {\sqrt {\map \cos {\theta_0 / 2} - \map \cos {\theta / 2} } } \rd \theta$
+Now we make the substitution:
+{{begin-eqn}}
+{{eqn | l = u
+      | r = \frac {\map \cos {\theta / 2} } {\map \cos {\theta_0 / 2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d u} {\d \theta}
+      | r = -\frac {\map \sin {\theta / 2} } {2 \map \cos {\theta_0 / 2} }
+      | c = 
+}}
+{{end-eqn}}
+Recalculating the limits:
+:when $\theta = \theta_0$ we have $u = 1$
+:when $\theta = \pi$ we have $u = 0$.
+So:
+{{begin-eqn}}
+{{eqn | l = T
+      | r = -2 \sqrt {\frac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }
+      | c = 
+}}
+{{eqn | r = \intlimits {2 \sqrt {\frac a g} \sin^{-1} u} 0 1
+      | c = 
+}}
+{{eqn | r = \pi \sqrt {\frac a g}
+      | c = 
+}}
+{{end-eqn}}
+Thus the [[Definition:Time|time]] to slide down a [[Definition:Brachistochrone|brachistochrone]] from any arbitrary point $\theta_0$ is:
+:$T = \pi \sqrt {\dfrac a g}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rational Numbers are F-Sigma Set in Real Line}
+Tags: F-Sigma Sets
+
+\begin{theorem}
+Let $\left({\R, \tau}\right)$ be the [[Definition:Real Number Line|real number line]] considered asa  [[Definition:Topological Space|topological space]] with the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
+Then:
+:$\Q$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]] in $\left({\R, \tau}\right)$.
+\end{theorem}
+
+\begin{proof}
+Define the [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $\R$ as:
+:$\mathcal F := \left\{{\left\{{x}\right\}: x \in \Q}\right\}$
+By [[Closed Real Interval is Closed Set]]:
+:$\forall x \in \Q: \left[{x \,.\,.\, x}\right] = \left\{{x}\right\}$ is [[Definition:Closed Set (Topology)|closed (in topological sense)]]
+Then:
+:$\forall A \in \mathcal F: A$ is [[Definition:Closed Set (Topology)|closed]]
+By [[Cardinality of Set of Singletons]]:
+:$\left\vert{\mathcal F}\right\vert = \left\vert{\Q}\right\vert$
+where $\left\vert{\mathcal F}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $\mathcal F$.
+By [[Rational Numbers are Countably Infinite]]:
+:$\Q$ is [[Definition:Countable Set|countable]].
+Therefore by [[Set is Countable if Cardinality equals Cardinality of Countable Set]]:
+:$\mathcal F$ is [[Definition:Countable Set|countable]].
+By [[Union of Set of Singletons]]:
+:$\bigcup \mathcal F = \Q$
+Thus, by definition, $\Q$ is an [[Definition:F-Sigma Set|$F_\sigma$ set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Union of Set of Singletons}
+Tags: Set Union, Singletons
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $T = \set {\set x: x \in S}$ be the [[Definition:Set|set]] of all [[Definition:Singleton|singletons]] of [[Definition:Element|elements]] of $S$.
+Then:
+:$\ds \bigcup T = S$
+where $\ds \bigcup T$ denotes the [[Definition:Union of Set of Sets|union]] of $T$.
+\end{theorem}
+
+\begin{proof}
+=== Union of $T$ Subset $S$ ===
+Let $\ds x \in \bigcup T$.
+By definition of [[Definition:Union of Set of Sets|union]]:
+:$\exists A \in T: x \in A$
+By definition of $T$:
+:$\exists y \in S: A = \set y$
+Then by definition of [[Definition:Singleton|singleton]]:
+:$x = y$
+Thus $x \in S$.
+{{qed|lemma}}
+=== $S$ Subset Union of $T$ ===
+Let $x \in S$.
+By definition of $T$:
+:$\set x \in T$
+By [[Set is Subset of Union/Set of Sets|Set is Subset of Union/Set of Sets]]:
+:$\ds \set x \subseteq \bigcup T$
+By definition of [[Definition:Singleton|singleton]]:
+:$x \in \set x$
+Thus by definition of [[Definition:Subset|subset]]:
+:$\ds x \in \bigcup T$
+{{qed|lemma}}
+Thus by definition of [[Definition:Set Equality|set equality]]:
+:$\ds \bigcup T = S$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Confocal Conics are Self-Orthogonal}
+Tags: Orthogonal Trajectories, Conic Sections
+
+\begin{theorem}
+The [[Definition:Confocal Conics|confocal conics]] defined by:
+:$\quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
+forms a [[Definition:Orthogonal Trajectories|family of orthogonal trajectories]] which is [[Definition:Self-Orthogonal Trajectories|self-orthogonal]].
+:[[File:ConfocalConics.png|500px]]
+\end{theorem}
+
+\begin{proof}
+Consider:
+:$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
+From [[Equation of Confocal Ellipses/Formulation 2|Equation of Confocal Ellipses: Formulation 2]]:
+:$(1)$ defines an [[Definition:Ellipse|ellipse]] when $a^2 > c^2$.
+From [[Equation of Confocal Hyperbolas/Formulation 2|Equation of Confocal Hyperbolas: Formulation 2]]:
+:$(1)$ defines a [[Definition:Hyperbola|hyperbola]] when $a^2 < c^2$.
+Thus it is seen that $(1)$ is that of a [[Definition:Conic Section|conic section]].
+We use the technique of [[Definition:Formation of Ordinary Differential Equation by Elimination|formation of ordinary differential equation by elimination]].
+[[Definition:Differentiation|Differentiating]] {{WRT|Differentiation}} $x$ gives:
+:$\dfrac {2 x} {a^2} + \dfrac {2 y} {a^2 - c^2} \dfrac {\d y} {\d x} = 0$
+so
+:$\dfrac {\d y} {\d x} = - \dfrac {a^2 - c^2} {a^2} \dfrac x y$
+Now we need to eliminate $c$ from the above.
+We go back to $(1)$:
+{{begin-eqn}}
+{{eqn | l = \frac {x^2} {a^2} + \frac {y^2} {a^2 - c^2}
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y^2} {a^2 - c^2}
+      | r = 1 - \frac {x^2} {a^2}
+      | c = 
+}}
+{{eqn | r = \frac {a^2 - x^2}{a^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a^2 - c^2
+      | r = \frac {a^2 y^2} {a^2 - x^2}
+      | c = 
+}}
+{{end-eqn}}
+Substituting for $a^2 - c^2$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d x}
+      | r = -\frac {a^2 y^2} {a^2 \paren {a^2 - x^2} } \frac x y
+      | c = 
+}}
+{{eqn | r = \frac {x y} {a^2 - x^2}
+      | c = 
+}}
+{{end-eqn}}
+This is separable, so separate it:
+:$\displaystyle \int \frac {\d y} y = -\int \frac {x \rd x} {a^2 - x^2}$
+which leads to:
+{{begin-eqn}}
+{{eqn | l = \ln y
+      | r = \frac 1 2 \, \map \ln {a^2 - x^2} + \ln k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \ln y
+      | r = \map \ln {a^2 - x^2} + 2 \ln k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \ln y^2
+      | r = \map \ln {a^2 - x^2} + \ln k^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \ln y^2
+      | r = \map \ln {k^2 \paren {a^2 - x^2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^2
+      | r = k^2 \paren {a^2 - x^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^2 + k^2 x^2
+      | r = a^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x^2} {a^2} + \frac {y^2} {k^2 a^2}
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Now $k^2$ is still arbitrary at this point, so set:
+:$k^2 = \dfrac {a^2 - c^2} {a^2}$
+Substituting this into the above gives us:
+:$\dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
+which is $(1)$.
+Hence the result by definition of [[Definition:Self-Orthogonal Trajectories|self-orthogonal]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T1 Space is T1/2 Space}
+Tags: T1 Spaces, T1: 2 Spaces
+
+\begin{theorem}
+Let $T$ be a [[Definition:T1 Space|$T_1$]] [[Definition:Topological Space|topological space]].
+Then $T$ is [[Definition:T1/2 Space|$T_{\frac 1 2}$ space]].
+\end{theorem}
+
+\begin{proof}
+By [[Closure of Derivative is Derivative in T1 Space]]:
+:$\forall A \subseteq T: \left({A'}\right)^- = A'$
+where
+:$A'$ denotes the [[Definition:Set Derivative|derivative]] of $A$
+:$\left({A'}\right)^-$ denotes the [[Definition:Closure (Topology)|closure]] of $A'$
+Then by [[Topological Closure is Closed]]:
+:$\forall A \subseteq T: A'$ is [[Definition:Closed Set (Topology)|closed]]
+Thus by definition:
+:$T$ is [[Definition:T1/2 Space|$T_{\frac 1 2}$ space]]
+{{qed}}
+\end{proof}<|endoftext|>
+\section{T1/2 Space is T0 Space}
+Tags: T0 Spaces, T1: 2 Spaces
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:T1/2 Space|$T_{\frac 1 2}$ topological space]].
+Then $T$ is [[Definition:T0 Space|$T_0$ space]].
+\end{theorem}
+
+\begin{proof}
+By [[Characterization of T0 Space by Closures of Singletons]] it suffices to prove that
+:$\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$
+where $\set y^-$ denotes the [[Definition:Closure (Topology)|closure]] of $\set y$.
+Let $x, y$ be [[Definition:Element|points]] of $T$ such that:
+:$x \ne y$
+{{AimForCont}}:
+:$x \in \set y^- \land y \in \set x^-$
+We will prove that:
+:$x \notin \set x'$
+where $\set x'$ denotes the [[Definition:Set Derivative|derivative]] of $\set x$.
+{{AimForCont}}:
+:$x \in \set x'$
+As $S$ is [[Definition:Open Set (Topology)|open]] by definition of [[Definition:Topological Space|topological space]]
+by [[Characterization of Derivative by Open Sets]]:
+:$\exists z \in S: z \in \set x \cap S \land z \ne x$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$z \in \set x$
+This contradicts $z \ne x$ by definition of [[Definition:Singleton|singleton]].
+Thus $x \notin \set x'$.
+We will prove that:
+:$(1): \quad \lnot \forall G \in \tau: y \in G \implies \set x \cap G \ne \O$
+{{AimForCont}}:
+:$\forall G \in \tau: y \in G \implies \set x \cap G \ne \O$
+As sublemma we will show that:
+:$\forall U \in \tau: y \in U \implies \exists r \in S: r \in \set x \cap U \land y \ne r$
+Let $U \in \tau$ such that:
+:$y \in U$
+Then by assumption:
+:$\set x \cap U \ne \O$
+By definition of [[Definition:Empty Set|empty set]]:
+:$\exists z: z \in \set x \cap U$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$z \in \set x$
+Then by definition of [[Definition:Singleton|singleton]]:
+:$z = x \ne y$
+Thus:
+:$\exists r \in S: r \in \set x \cap U \land y \ne r$
+Then by [[Characterization of Derivative by Open Sets]]:
+:$y \in \set x'$
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$y \notin \relcomp S {\set x'} \land x \in \relcomp S {\set x'}$
+By definition of [[Definition:T1/2 Space|$T_{\frac 1 2}$ space]]:
+:$\set x'$ is [[Definition:Closed Set (Topology)|closed]]
+By definition of [[Definition:Closed Set (Topology)|closed set]]:
+:$\relcomp S {\set x'}$ is [[Definition:Open Set (Topology)|open]]
+By $x \in \set y^-$ and by [[Condition for Point being in Closure]]:
+:$\set y \cap \relcomp S {\set x'} \ne \O$
+Then by definition of [[Definition:Empty Set|empty set]]:
+:$\exists z: z \in \set y \cap \relcomp S {\set x'}$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$z \in \set y \land z \in \relcomp S {\set x'}$
+By definition of [[Definition:Singleton|singleton]]:
+:$z = y$
+Thus $z \in \relcomp S {\set x'}$ contradicts $y \notin \relcomp S {\set x'}$.
+This ends the proof of $(1)$.
+Then by $(1)$ and [[Condition for Point being in Closure]]:
+:$y \notin \set x^-$
+This contradicts $y \in \set x^-$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Parabolas Inscribed in Shared Tangent Lines}
+Tags: Analytic Geometry
+
+\begin{theorem}
+Let the function $\map f x = A x^2 + B x + C_1$ be a curve embedded in the [[Definition:Euclidean Plane|Euclidean Plane]].
+Let $\map {y_1} x$ be the equation of the [[Definition:Tangent to Curve|tangent]] line at $\tuple {Q, \map f Q}$ on $f$.
+Let $\map {y_2} x$ be the equation of the [[Definition:Tangent to Curve|tangent]] line at $\tuple {-Q, \map f {-Q} }$ on $f$.
+Then there exists another function $\map g x$ also embedded in the [[Definition:Euclidean Plane|Euclidean Plane]] defined as:
+:$\map g x = -A x^2 + B x + C_2$.
+with:
+:tangent lines $\map {y_3} x$ being the equation of the [[Definition:Tangent to Curve|tangent]] line at $\tuple {Q, \map g Q}$ on $g$
+and:
+:$\map {y_4} x$ being the equation of the [[Definition:Tangent to Curve|tangent]] line at $\tuple {-Q, \map g {-Q} }$ on $g$.
+so that the [[Definition:Tangent to Curve|tangent]] lines $y_3$ and $y_4$ inscribe $\map f x$ and the [[Definition:Tangent to Curve|tangent]] lines $y_1$ and $y_2$ inscribe $\map g x$.
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Tangent to Curve|tangent]] line at $\tuple {Q, \map f \Q}$ on $f$ is defined as:
+:$\map {y_1} x = \paren {2 A Q + B} x + b_1$
+where $2 A Q + B$ is the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] line on the point $\tuple {Q, \map g Q}$ on $f$.
+  
+Substitute in the [[Definition:Coordinate System|coordinates]] of the point $\tuple {Q, \map g Q}$ to $y_1$ and solve for $b_1$.
+This will reveal the $y$-intercept of $y_1$:
+{{begin-eqn}}
+{{eqn| l = A Q^2 + B Q + C_1
+     | r = \paren {2 A Q + B} Q + b_1
+     | c = The value of $y_1 = \map f Q$
+}}
+{{eqn| l = A Q^2 + B Q + C_1 -2 A Q^2 - B Q
+     | r = b_1
+}}
+{{eqn| l = -A Q^2 + C_1
+     | r = b_1
+}}
+{{end-eqn}}
+    
+Continue by following the same steps for $y_2$ which is defined:
+:$\map {y_2} x = \paren {-2 A Q + B} x + b_2$
+where $-2 A Q + B$ is the slope of the [[Definition:Tangent to Curve|Tangent]] line at the point $\tuple {-Q, \map f {-Q} }$ on $f$.
+Substitute in the [[Definition:Coordinate System|coordinates]] of the point $\paren {-Q, \map f {-Q} }$ to $y_2$.
+Use these values to solve for $b_2$, and this will reveal the $y$-intercept of $y_2$:
+{{begin-eqn}}
+{{eqn | l = A \paren {-Q}^2 + B \paren {-Q} + C_1
+      | r = \paren {-2 A Q + B} \paren {-Q} + b_2
+      | c = the value of $y_2 = \map f {-Q}$
+}}
+{{eqn | l = A Q^2 - B Q + C_1 -2 A Q^2 + B Q
+      | r = b_2
+}}
+{{eqn | l = -A Q^2 + C_1
+      | r = b_2
+}}
+{{end-eqn}}
+The $y$-intercepts of both $y_1$ and $y_2$ have been shown to be equivalent.
+{{qed|lemma}}
+Since $b_1 = b_2$ redefine this value as $b$.
+The [[Definition:Distance Between Points|distance]] between $b$ and $C_1$ is $\size {C_1 - b}$.
+Let $\map g x = -A x^2 + B x + C_2$.
+Then the [[Definition:Tangent to Curve|Tangent]] line at the point $\tuple {Q, \map g Q}$ on $g$ is defined as:
+:$\map {y_3} x = \paren {-2 A Q + B} x + b_3$
+where $-2 A Q + B$ is the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] line at $\tuple {Q, \map g Q}$ on $g$.
+Solve for $b_3$ using the same methods used for $y_1$ and $y_2$.
+This will reveal the $y$-intercept of $y_3$:
+:$b_3 = A Q^2 + C_2$
+The result also follows for the [[Definition:Tangent to Curve|Tangent]] line $\tuple {-Q, \map g {-Q} }$ on $g$ which is defined:
+:$y_4 = \paren {-2 A Q + B} x + b_4$
+Solving for $b_4$ yields the result:
+:$b_4 = A Q^2 + C_2$
+The $y$-intercepts of both $y_3$ and $y_4$ have been shown to be equivalent.
+{{qed|lemma}}
+Notice that the derivatives of $f$ and $g$ satisfy:
+{{begin-eqn}}
+{{eqn | l = \map {g'} Q
+      | r = \map {f'} {-Q}
+}}
+{{eqn | l = \map {g'} {-Q}
+      | r = \map {f'} Q
+}}
+{{end-eqn}}
+Then it must be true that:
+:$y_1 = y_4$ and $y_2 = y_3$
+and the functions $y_1$, $y_2$, $y_3$, and $y_4$ share the same $y$-intercept.
+{{qed|lemma}}
+Redefine this the $y$-intercepts of the tangent lines as $b$.
+Solve for $C_2$ to determine the vertical translation of $\map g x$:
+{{begin-eqn}}
+{{eqn | l = C_2
+      | r = \paren {-A Q^2 + C_1} - A Q^2
+}}
+{{eqn | l = C_2
+      | r = -2 A Q^2 + C_1
+}}
+{{end-eqn}}
+Therefore the function:
+:$\map g x = -A x^2 + B x - \paren {2 A Q^2 + C_1}$
+will have [[Definition:Tangent to Curve|tangent]] lines equivalent to the [[Definition:Tangent to Curve|tangent]] lines on $\map f x$ at the points $\tuple {Q, \map f Q}$, and $\tuple {-Q, \map f {-Q} }$.
+{{qed}}
+[[Category:Analytic Geometry]]
+ao18o8hiwqw6up7edqzndcsegp1i1r5
+\end{proof}<|endoftext|>
+\section{Characterization of T0 Space by Closures of Singletons}
+Tags: T0 Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Then
+:$T$ is a [[Definition:T0 Space|$T_0$ space]] {{iff}}
+::$\forall x, y \in S: x \ne y \implies x \notin \left\{{y}\right\}^- \lor y \notin \left\{{x}\right\}^-$
+where $\left\{{y}\right\}^-$ denotes the [[Definition:Closure (Topology)|closure]] of $\left\{{y}\right\}$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $T$ be a [[Definition:T0 Space|$T_0$ space]].
+Let $x, y \in S$ such that
+:$x \ne y$
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$
+Then:
+:$\left\{{x}\right\} \subseteq \left\{{y}\right\}^- \land \left\{{y}\right\} \subseteq \left\{{x}\right\}^-$
+By [[Topological Closure of Subset is Subset of Topological Closure]]:
+:$\left\{{x}\right\}^- \subseteq \left({\left\{{y}\right\}^-}\right)^- \land \left\{{y}\right\}^- \subseteq \left({\left\{{x}\right\}^-}\right)^-$
+By [[Closure of Topological Closure equals Closure]]:
+:$\left\{{x}\right\}^- \subseteq \left\{{y}\right\}^- \land \left\{{y}\right\}^- \subseteq \left\{{x}\right\}^-$
+Then by definition of [[Definition:Set Equality|set equality]]:
+:$\left\{{x}\right\}^- = \left\{{y}\right\}^-$
+By [[Characterization of T0 Space by Distinct Closures of Singletons|Characterization of $T_0$ Space by Distinct Closures of Singletons]]:
+:$\left\{{x}\right\}^- \ne \left\{{y}\right\}^-$
+This contradicts the equality.
+Thus the result by [[Proof by Contradiction]]
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that:
+:$(1): \quad \forall x, y \in S: x \ne y \implies x \notin \left\{{y}\right\}^- \lor y \notin \left\{{x}\right\}^-$
+By [[Characterization of T0 Space by Distinct Closures of Singletons|Characterization of $T_0$ Space by Distinct Closures of Singletons]] it suffices to prove
+:$\forall x, y \in S: x \ne y \implies \left\{{y}\right\}^- \ne \left\{{x}\right\}^-$
+Let $x, y \in S$ such that
+:$x \ne y$
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$(2): \quad \left\{{y}\right\}^- = \left\{{x}\right\}^-$
+By definition of [[Definition:Singleton|singleton]]:
+:$x \in \left\{{x}\right\} \land y \in \left\{{y}\right\}$
+By [[Set is Subset of its Topological Closure]]
+:$\left\{{x}\right\} \subseteq \left\{{x}\right\}^- \land \left\{{y}\right\} \subseteq \left\{{y}\right\}^-$
+Then by definition of [[Definition:Subset|subset]] and $(2)$:
+:$x \in \left\{{y}\right\}^- \land y \in \left\{{x}\right\}^-$
+This contradicts the assumption $(1)$.
+Thus the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of T0 Space by Distinct Closures of Singletons}
+Tags: T0 Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Then
+:$T$ is a [[Definition:T0 Space|$T_0$ space]] {{iff}}
+::$\forall x, y \in S: x \ne y \implies \left\{{x}\right\}^- \ne \left\{{y}\right\}^-$
+where $\left\{{y}\right\}^-$ denotes the [[Definition:Closure (Topology)|closure]] of $\left\{{y}\right\}$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $T$ be a [[Definition:T0 Space|$T_0$ space]].
+Let $x, y \in S$ such that
+:$x \ne y$
+By definition of [[Definition:T0 Space|$T_0$ space]]:
+:$\left({\exists U \in \tau: x \in U \land y \notin U}\right) \lor \left({\exists U \in \tau: x \notin U \land y \in U}\right)$
+[[Definition:WLOG|WLOG]]: Suppose
+:$\exists U \in \tau: x \in U \land y \notin U$
+By definition of [[Definition:Singleton|singleton]]:
+:$x \in \left\{{x}\right\}$
+By [[Set is Subset of its Topological Closure]]:
+:$\left\{{x}\right\} \subseteq \left\{{x}\right\}^-$
+Then by definition of [[Definition:Subset|subset]]:
+:$x \in \left\{{x}\right\}^-$
+Hence by definition of [[Definition:Set Intersection|intersection]]:
+:$\left\{{x}\right\}^- \cap U \ne \varnothing$
+By definition of [[Definition:Singleton|singleton]]:
+:$\forall z: z \in \left\{{y}\right\} \implies z = y$
+Then by definition of [[Definition:Set Intersection|intersection]]:
+:$\left\{{y}\right\} \cap U = \varnothing$
+Hence by [[Disjoint Open Sets remain Disjoint with one Closure]]
+:$\left\{{y}\right\}^- \cap U = \varnothing$
+Thus:
+:$\left\{{x}\right\}^- \ne \left\{{y}\right\}^-$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that:
+:$(1): \quad \forall x, y \in S: x \ne y \implies \left\{{x}\right\}^- \ne \left\{{y}\right\}^-$
+By [[Characterization of T0 Space by Closed Sets|Characterization of $T_0$ Space by Closed Sets]] it suffices to prove that
+:$\left({\exists F \subseteq S: F}\right.$ is [[Definition:Closed Set (Topology)|closed]] $\left. {\land x \in F \land y \notin F}\right)$
+or
+:$\left({\exists F \subseteq S: F}\right.$ is [[Definition:Closed Set (Topology)|closed]] $\left. {\land x \notin F \land y \in F}\right)$
+Then assume:
+:$(2): \quad \forall F \subseteq S: F$ is [[Definition:Closed Set (Topology)|closed]] $\land x \in F \implies y \in F$
+Define $F := \left\{{y}\right\}^-$
+By [[Topological Closure is Closed]]:
+:$F$ is [[Definition:Closed Set (Topology)|closed]].
+We will prove now that
+:$x \notin F$
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$x \in F$
+Then:
+:$\left\{{x}\right\} \subseteq F$
+By [[Topological Closure of Subset is Subset of Topological Closure]]:
+:$\left\{{x}\right\}^- \subseteq F^-$
+By [[Set is Closed iff Equals Topological Closure]]:
+:$(3): \quad \left\{{x}\right\}^- \subseteq F$
+By definition of [[Definition:Singleton|singleton]]:
+:$x \in \left\{{x}\right\}$
+By [[Set is Subset of its Topological Closure]]:
+:$\left\{{x}\right\} \subseteq \left\{{x}\right\}^-$
+Then by definition of [[Definition:Subset|subset]]:
+:$x \in \left\{{x}\right\}^-$
+Because $\left\{{x}\right\}^-$ is [[Definition:Closed Set (Topology)|closed]] therefore by $(2)$:
+:$y \in \left\{{x}\right\}^-$
+Then:
+:$\left\{{y}\right\} \subseteq \left\{{x}\right\}^-$
+By [[Topological Closure of Subset is Subset of Topological Closure]]:
+:$F \subseteq \left({\left\{{x}\right\}^-}\right)^-$
+Then by [[Closure of Topological Closure equals Closure]]:
+:$F \subseteq \left\{{x}\right\}^-$
+Hence by $(3)$ and definition of [[Definition:Set Equality|set equality]]:
+:$F = \left\{{x}\right\}^-$
+This contradicts the assumption $(1)$.
+Thus $x \notin F$.
+By definition of [[Definition:Singleton|singleton]]:
+:$y \in \left\{{y}\right\}$
+By [[Set is Subset of its Topological Closure]]:
+:$\left\{{y}\right\} \subseteq \left\{{y}\right\}^- = F$
+Thus by definition of [[Definition:Subset|subset]]:
+:$y \in F$
+Hence
+:$\exists F \subseteq S: F$ is [[Definition:Closed Set (Topology)|closed]] $\land x \notin F \land y \in F$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Characterization of T0 Space by Closed Sets}
+Tags: T0 Spaces
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Then
+:$T$ is a [[Definition:T0 Space|$T_0$ space]] {{iff}}
+::for every [[Definition:Element|points]] $x, y \in S$ if $x \ne y$ then
+:::there exists a [[Definition:Closed Set (Topology)|closed]] [[Definition:Subset|subset]] $F$ of $S$ such that $x \in F$ and $y \notin F$
+::or
+:::there exists a [[Definition:Closed Set (Topology)|closed]] [[Definition:Subset|subset]] $F$ of $S$ such that $x \notin F$ and $y \in F$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $T$ be a [[Definition:T0 Space|$T_0$ space]].
+Let $x, y \in S$ such that
+:$x \ne y$
+By definition of [[Definition:T0 Space|$T_0$ space]]:
+:$\left({\exists U \in \tau: x \in U \land y \notin U}\right) \lor \left({\exists U \in \tau: x \notin U \land y \in U}\right)$
+[[Definition:WLOG|WLOG]]: Suppose:
+:$\exists U \in \tau: x \in U \land y \notin U$
+By definition:
+:$\complement_S\left({U}\right)$ is [[Definition:Closed Set (Topology)|closed]]
+where $\complement_S\left({U}\right)$ denotes the [[Definition:Relative Complement|relative complement]] of $U$ in $S$.
+By definition of [[Definition:Relative Complement|relative complement]]:
+:$x \notin \complement_S\left({U}\right) \land y \in \complement_S\left({U}\right)$
+Thus:
+:$\exists F \subseteq S: F$ is [[Definition:Closed Set (Topology)|closed]] $\land\, x \notin F \land y \in F$
+{{qed|lemma}}
+=== Necessary Condition ===
+This statement follows [[Definition:Mutatis Mutandis|mutatis mutandis]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Focus of Ellipse from Major and Minor Axis}
+Tags: Ellipses
+
+\begin{theorem}
+Let $K$ be an [[Definition:Ellipse|ellipse]] whose [[Definition:Major Axis of Ellipse|major axis]] is $2 a$ and whose [[Definition:Minor Axis of Ellipse|minor axis]] is $2 b$.
+Let $c$ be the [[Definition:Distance (Linear Measure)|distance]] of the [[Definition:Focus of Ellipse|foci]] of $K$ from the [[Definition:Center of Ellipse|center]].
+Then:
+:$a^2 = b^2 + c^2$
+\end{theorem}
+
+\begin{proof}
+:[[File:EllipseFocus MajorMinorAxes.png|500px]]
+Let the [[Definition:Focus of Ellipse|foci]] of $K$ be $F_1$ and $F_2$.
+Let the [[Definition:Vertex of Ellipse|vertices]] of $K$ be $V_1$ and $V_2$.
+Let the [[Definition:Covertex of Ellipse|covertices]] of $K$ be $C_1$ and $C_2$.
+Let $P = \left({x, y}\right)$ be an arbitrary [[Definition:Point|point]] on the [[Definition:Locus|locus]] of $K$.
+From the [[Definition:Equidistance Property of Ellipse|equidistance property]] of $K$ we have that:
+:$F_1 P + F_2 P = d$
+where $d$ is a [[Definition:Constant|constant]] for this particular [[Definition:Ellipse|ellipse]].
+This is true for all [[Definition:Point|points]] on $K$.
+In particular, it holds true for $V_2$, for example.
+Thus:
+{{begin-eqn}}
+{{eqn | l = d
+      | r = F_1 V_2 + F_2 V_2
+      | c = 
+}}
+{{eqn | r = \paren {a + c} + \paren {a - c}
+      | c = 
+}}
+{{eqn | r = 2 a
+      | c = 
+}}
+{{end-eqn}}
+It also holds true for $C_2$:
+:$F_1 C_2 + F_2 C_2 = d$
+Then:
+{{begin-eqn}}
+{{eqn | l = F_1 C_2^2
+      | r = O F_1^2 + O C_2^2
+      | c = [[Pythagoras's Theorem]]
+}}
+{{eqn | r = c^2 + b^2
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = F_1 C_2^2
+      | r = O F_1^2 + O C_2^2
+      | c = [[Pythagoras's Theorem]]
+}}
+{{eqn | r = c^2 + b^2
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = F_1 C_2 + F_2 C_2
+      | r = 2 \sqrt {b^2 + c^2}
+      | c = 
+}}
+{{eqn | r = 2 a
+      | c = as $2 a = d$
+}}
+{{eqn | ll= \leadsto
+      | l = a
+      | r = \sqrt {b^2 + c^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | r = a^2
+      | l = b^2 + c^2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equidistance of Ellipse equals Major Axis}
+Tags: Ellipses
+
+\begin{theorem}
+Let $K$ be an [[Definition:Ellipse|ellipse]] whose [[Definition:Focus of Ellipse|foci]] are $F_1$ and $F_2$.
+Let $P$ be an arbitrary [[Definition:Point|point]] on $K$.
+Let $d$ be the [[Definition:Constant|constant]] [[Definition:Distance (Linear Measure)|distance]] such that:
+:$d_1 + d_2 = d$
+where:
+: $d_1 = P F_1$
+: $d_2 = P F_2$
+Then $d$ is equal to the [[Definition:Major Axis of Ellipse|major axis]] of $K$.
+\end{theorem}
+
+\begin{proof}
+:[[File:EllipseEquidistanceMajorAxis.png|400px]]
+By the [[Definition:Equidistance Property of Ellipse|equidistance property]] of $K$:
+:$d_1 + d_2 = d$
+applies to all [[Definition:Point|points]] $P$ on $K$.
+Thus it also applies to the two [[Definition:Vertex of Ellipse|vertices]] $V_1$ and $V_2$:
+:$V_1 F_1 + V_1 F_2 = d$
+:$V_2 F_1 + V_2 F_2 = d$
+Adding:
+:$V_1 F_1 + V_2 F_1 + V_1 F_2 + V_2 F_2 = 2 d$
+But:
+:$V_1 F_1 + V_2 F_1 = V_1 V_2$
+:$V_1 F_2 + V_2 F_2 = V_1 V_2$
+and so:
+:$2 V_1 V_2 = 2 d$
+By definition, the [[Definition:Major Axis of Ellipse|major axis]] is $V_1 V_2$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Equidistance of Hyperbola equals Transverse Axis}
+Tags: Hyperbolas
+
+\begin{theorem}
+Let $K$ be an [[Definition:Hyperbola|hyperbola]] whose [[Definition:Focus of Hyperbola|foci]] are $F_1$ and $F_2$.
+Let $P$ be an arbitrary [[Definition:Point|point]] on $K$.
+Let $d$ be the [[Definition:Constant|constant]] [[Definition:Distance (Linear Measure)|distance]] such that:
+:$\left\lvert{d_1 - d_2}\right\rvert = d$
+where:
+: $d_1 = P F_1$
+: $d_2 = P F_2$
+Then $d$ is equal to the [[Definition:Transverse Axis of Hyperbola|transverse axis]] of $K$.
+\end{theorem}
+
+\begin{proof}
+&nbsp;
+:[[File:HyperbolaEquidistanceTransverseAxis.png|400px]]
+By the [[Definition:Equidistance Property of Hyperbola|equidistance property]] of $K$:
+:$\left\lvert{d_1 - d_2}\right\rvert = d$
+applies to all [[Definition:Point|points]] $P$ on $K$.
+Thus it also applies to the two [[Definition:Vertex of Hyperbola|vertices]] $V_1$ and $V_2$.
+Observing the signs of $\left\lvert{d_1 - d_2}\right\rvert$ as appropriate:
+:$V_1 F_2 - V_1 F_1 = d$
+:$V_2 F_1 - V_2 F_2 = d$
+Adding:
+:$\left({V_1 F_2 - V_2 F_2}\right) + \left({V_2 F_1 - V_1 F_1}\right) = 2 d$
+But:
+:$V_1 F_2 - V_2 F_2 = V_1 V_2$
+:$V_2 F_1 - V_1 F_1 = V_1 V_2$
+and so:
+:$2 V_1 V_2 = 2 d$
+By definition, the [[Definition:Transverse Axis of Hyperbola|transverse axis]] of $K$ is $V_1 V_2$.
+Hence the result.
+{{qed}}
+[[Category:Hyperbolas]]
+slpd59kn99njqsfae0azz17ox2rwikm
+\end{proof}<|endoftext|>
+\section{Equivalence of Definitions of Ellipse}
+Tags: Ellipses
+
+\begin{theorem}
+The following definitions of an [[Definition:Ellipse|ellipse]] are [[Definition:Logical Equivalence|equivalent]]:
+\end{theorem}
+
+\begin{proof}
+Let $K$ be an [[Definition:Ellipse|ellipse]] aligned in a [[Definition:Cartesian Plane|cartesian plane]] in [[Definition:Reduced Form of Ellipse|reduced form]].
+Thus its [[Definition:Focus of Ellipse|foci]] are at $\tuple {\mathop \pm c, 0}$.
+Let:
+:the [[Definition:Major Axis of Ellipse|major axis]] of $K$ have [[Definition:Length (Linear Measure)|length]] $2 a$
+:the [[Definition:Minor Axis of Ellipse|minor axis]] of $K$ have [[Definition:Length (Linear Measure)|length]] $2 b$.
+From [[Equation of Ellipse in Reduced Form/Cartesian Frame|Equation of Ellipse in Reduced Form]], the equation of $K$ is:
+:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
+This has been derived from the [[Definition:Equidistance Property of Ellipse|equidistance property]] of $K$.
+From [[Focus of Ellipse from Major and Minor Axis]]:
+:$a^2 - c^2 = b^2$
+and so $K$ can be expressed as:
+:$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
+It remains to be shown that $K$ possesses the [[Definition:Ellipse/Focus-Directrix|focus-directrix property]].
+:[[File:EllipseFocusDirectrixEquidistance.png|400px]]
+Let $D$ be the [[Definition:Straight Line|straight line]] whose equation is $x = \dfrac {a^2} c$.
+It will be shown that the [[Definition:Locus|locus]] of [[Definition:Point|points]] $P = \tuple {x, y}$ such that:
+:the [[Definition:Distance (Linear Measure)|distance]] from $P$ to $F_1$ is $\dfrac c a$ of the [[Definition:Distance (Linear Measure)|distance]] from $P$ to $D$
+is precisely equation $(1)$.
+We have that:
+:$P F_2 = \epsilon \paren {d - x}$
+where:
+:$\epsilon = \dfrac c a$
+:$d = \dfrac {a^2} c$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \sqrt {\paren {x - c}^2 + y^2}
+      | r = \frac c a \paren {\frac {a^2} c - x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x - c}^2 + y^2
+      | r = \paren {a - \frac c a x}^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 - 2 c x + c^2 + y^2
+      | r = a^2 - 2 c x + \frac {c^2} {a^2} x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 \paren {1 - \frac {c^2} {a^2} } + y^2
+      | r = a^2 - c^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a^2 - c^2} \frac {x^2} {a^2} + y^2
+      | r = a^2 - c^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x^2} {a^2} + \frac {y^2} {a^2 - c^2}
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Ellipses]]
+5qzqtap264tdqssj29kxqz3bdgmv01k
+\end{proof}<|endoftext|>
+\section{Equation of Hyperbola in Reduced Form}
+Tags: Hyperbolas
+
+\begin{theorem}
+Let $K$ be an [[Definition:Hyperbola|hyperbola]] aligned in a [[Definition:Cartesian Plane|cartesian plane]] in [[Definition:Reduced Form of Hyperbola|reduced form]].
+Let:
+:the [[Definition:Transverse Axis of Hyperbola|transverse axis]] of $K$ have [[Definition:Length (Linear Measure)|length]] $2 a$
+:the [[Definition:Conjugate Axis of Hyperbola|conjugate axis]] of $K$ have [[Definition:Length (Linear Measure)|length]] $2 b$.
+\end{theorem}<|endoftext|>
+\section{Equation of Hyperbola in Reduced Form/Cartesian Frame}
+Tags: Hyperbolas
+
+\begin{theorem}
+The equation of $K$ is:
+:$\dfrac {x^2} {a^2} - \dfrac {y^2} {b^2} = 1$
+\end{theorem}
+
+\begin{proof}
+&nbsp;
+:[[File:HyperbolaReducedForm.png|500px]]
+By definition, the [[Definition:Focus of Hyperbola|foci]] $F_1$ and $F_2$ of $K$ are located at $\tuple {-c, 0}$ and $\tuple {c, 0}$ respectively.
+Let the [[Definition:Vertex of Hyperbola|vertices]] of $K$ be $V_1$ and $V_2$.
+By definition, these are located at $\tuple {-a, 0}$ and $\tuple {a, 0}$.
+Let the [[Definition:Covertex of Hyperbola|covertices]] of $K$ be $C_1$ and $C_2$.
+By definition, these are located at $\tuple {0, -b}$ and $\tuple {0, b}$.
+Let $P = \tuple {x, y}$ be an arbitrary [[Definition:Point|point]] on the [[Definition:Locus|locus]] of $K$.
+From the [[Definition:Equidistance Property of Hyperbola|equidistance property]] of $K$ we have that:
+:$\size {F_1 P - F_2 P} = d$
+where $d$ is a [[Definition:Constant|constant]] for this particular [[Definition:Ellipse|ellipse]].
+From [[Equidistance of Hyperbola equals Transverse Axis]]:
+:$d = 2 a$
+Also, from [[Focus of Hyperbola from Transverse and Conjugate Axis]]:
+:$c^2 a^2 = b^2$
+{{WLOG}}, let us choose a [[Definition:Point|point]] $P$ such that $F_1 P > F_2 P$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \sqrt {\paren {x + c}^2 + y^2} - \sqrt {\paren {x - c}^2 + y^2}
+      | r = d = 2 a
+      | c = [[Pythagoras's Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = \sqrt {\paren {x + c}^2 + y^2}
+      | r = 2 a + \sqrt {\paren {x - c}^2 + y^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x + c}^2 + y^2
+      | r = \paren {2 a + \sqrt {\paren {x - c}^2 + y^2} }^2
+      | c = [[Definition:Squaring|squaring]] both sides
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 + 2 c x + c^2 + y^2
+      | r = 4 a^2 + 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2
+      | c = expanding
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 + 2 c x + c^2 + y^2
+      | r = 4 a^2 + 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2
+      | c = further expanding
+}}
+{{eqn | ll= \leadsto
+      | l = c x - a^2
+      | r = a \sqrt {\paren {x - c}^2 + y^2}
+      | c = gathering terms and simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {c x - a^2}^2
+      | r = a^2 \paren {\paren {x - c}^2 + y^2}^2
+      | c = [[Definition:Squaring|squaring]] both sides
+}}
+{{eqn | ll= \leadsto
+      | l = c^2 x^2 - 2 c x a^2 + a^4
+      | r = a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2
+      | c = expanding
+}}
+{{eqn | ll= \leadsto
+      | l = c^2 x^2 + a^4
+      | r = a^2 x^2 + a^2 c^2 + a^2 y^2
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = a^2 c^2 - a^4
+      | r = c^2 x^2 - a^2 x^2 - a^2 y^2
+      | c = gathering terms
+}}
+{{eqn | ll= \leadsto
+      | l = a^2 \paren {c^2 - a^2}
+      | r = \paren {c^2 - a^2} x^2 - a^2 y^2
+      | c = simplifying
+}}
+{{eqn | ll= \leadsto
+      | l = a^2 b^2
+      | r = b^2 x^2 - a^2 y^2
+      | c = substituting $c^2 - a^2 = b^2$ from $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = 1
+      | r = \frac {x^2} {a^2} - \frac {y^2} {b^2}
+      | c = dividing by $a^2 b^2$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Focus of Hyperbola from Transverse and Conjugate Axis}
+Tags: Hyperbolas
+
+\begin{theorem}
+Let $K$ be a [[Definition:Hyperbola|hyperbola]] whose [[Definition:Transverse Axis of Hyperbola|transverse axis]] is $2 a$ and whose [[Definition:Conjugate Axis of Hyperbola|conjugate axis]] is $2 b$.
+Let $c$ be the [[Definition:Distance (Linear Measure)|distance]] of the [[Definition:Focus of Hyperbola|foci]] of $K$ from the [[Definition:Center of Ellipse|center]].
+Then:
+:$c^2 = a^2 + b^2$
+\end{theorem}
+
+\begin{proof}
+&nbsp;
+:[[File:HyperbolaFocusTransConj.png|500px]]
+Let the [[Definition:Focus of Hyperbola|foci]] of $K$ be $F_1$ and $F_2$.
+Let the [[Definition:Vertex of Hyperbola|vertices]] of $K$ be $V_1$ and $V_2$.
+Let the [[Definition:Covertex of Hyperbola|covertices]] of $K$ be $C_1$ and $C_2$.
+Let $P = \tuple {x, y}$ be an arbitrary [[Definition:Point|point]] on the [[Definition:Locus|locus]] of $K$.
+From the [[Definition:Equidistance Property of Hyperbola|equidistance property]] of $K$ we have that:
+:$\size {F_1 P - F_2 P} = d$
+where $d$ is a [[Definition:Constant|constant]] for this particular [[Definition:Hyperbola|hyperbola]].
+This is true for all [[Definition:Point|points]] on $K$.
+In particular, it holds true for $V_2$, for example.
+Thus:
+{{begin-eqn}}
+{{eqn | l = d
+      | r = F_1 V_2 - F_2 V_2
+      | c = 
+}}
+{{eqn | r = \paren {c + a} - \paren {c - a}
+      | c = 
+}}
+{{eqn | r = 2 a
+      | c = 
+}}
+{{end-eqn}}
+{{finish|Some weird magic happens, and then:}}
+{{begin-eqn}}
+{{eqn | l = c
+      | r = \sqrt {a^2 + b^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | r = c^2
+      | l = a^2 + b^2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Hyperbolas]]
+5ciwgbtn56ylvhku9v7h5tt00apkaoi
+\end{proof}<|endoftext|>
+\section{Equation of Hyperbola in Reduced Form/Cartesian Frame/Parametric Form}
+Tags: Hyperbolas
+
+\begin{theorem}
+The equation of $K$ in [[Definition:Parametric Equation|parametric form]] is:
+:$x = a \cosh \theta, y = b \sinh \theta$
+\end{theorem}
+
+\begin{proof}
+Let the point $\tuple {x, y}$ satisfy the equations:
+:$x = a \cosh \theta$
+:$y = b \sinh \theta$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {x^2} {a^2} - \frac {y^2} {b^2}
+      | r = \frac {\paren {a \cosh \theta}^2} {a^2} - \frac {\paren {b \sinh \theta}^2} {b^2}
+      | c = 
+}}
+{{eqn | r =  \frac {a^2} {a^2} \cosh^2 \theta - \frac {b^2} {b^2} \sinh^2 \theta
+      | c = 
+}}
+{{eqn | r = \cosh^2 \theta - \sinh^2 \theta
+      | c = 
+}}
+{{eqn | r = 1
+      | c = [[Difference of Squares of Hyperbolic Cosine and Sine]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Condensation Points is Subset of Derivative}
+Tags: Condensation Points, Set Derivatives
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A$ be a [[Definition:Subset|subset]] of $S$.
+Then:
+:$A^0 \subseteq A'$
+where
+:$A^0$ denotes the [[Definition:Set of Condensation Points|set of condensation points]] of $A$
+:$A'$ denotes the [[Definition:Set Derivative|derivative]] of $A$
+\end{theorem}
+
+\begin{proof}
+Let $x \in A^0$.
+By definition of [[Definition:Set of Condensation Points|set of condensation points]]:
+:$x$ is [[Definition:Condensation Point|condensation point]] of $A$
+By definition of [[Definition:Condensation Point|condensation point]]:
+:$x$ is [[Definition:Limit Point of Set|limit point]] of $A$
+By definition of [[Definition:Derived Set|derived set]]:
+:$x \in A'$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closure of Set of Condensation Points equals Itself}
+Tags: Condensation Points, Set Closures
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A$ be a [[Definition:Subset|subset]] of $S$.
+Then:
+:$\left({A^0}\right)^- = A^0$
+where
+:$A^0$ denotes the [[Definition:Set of Condensation Points|set of condensation points]] of $A$
+:$A^-$ denotes the [[Definition:Closure (Topology)|closure]] of $A$
+\end{theorem}
+
+\begin{proof}
+By [[Set is Subset of its Topological Closure]]:
+:$A^0 \subseteq \left({A^0}\right)^-$
+To prove the equality by definition of [[Definition:Set Equality|set equality]] it suffices to show the inclusion:
+:$\left({A^0}\right)^- \subseteq A^0$
+Let $x \in \left({A^0}\right)^-$.
+We will prove that
+:$(1): \quad \forall U \in \tau: x \in U \implies A \cap U$ is [[Definition:Countable Set|uncountable]]
+Let $U$ be an [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subset]] of $S$ such that
+:$x \in U$
+By [[Condition for Point being in Closure]]:
+:$A^0 \cap U \ne \varnothing$
+By definition of [[Definition:Empty Set|empty set]]:
+:$\exists y: y \in A^0 \cap U$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$y \in A^0 \land y \in U$
+By definition of [[Definition:Set of Condensation Points|set of condensation points]]:
+:$y$ is [[Definition:Condensation Point|condensation point]] of $A$
+Then by definition of [[Definition:Condensation Point|condensation point]]:
+:$A \cap U$ is [[Definition:Countable Set|uncountable]]
+We will prove that
+:$x$ is [[Definition:Limit Point of Set|limit point]] of $A$
+Let $U$ be an [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subset]] of $S$ such that
+:$x \in U$
+By $(1)$:
+:$A \cap U$ is [[Definition:Countable Set|uncountable]]
+:$A \cap \left({U \setminus \left\{{x}\right\}}\right)$ is [[Definition:Countable Set|uncountable]]
+Thus
+:$A \cap \left({U \setminus \left\{{x}\right\}}\right) \ne \varnothing$
+Then by definition and $(1)$:
+:$x$ is [[Definition:Condensation Point|condensation point]] of $A$
+Thus by definition of [[Definition:Set of Condensation Points|set of condensation points]]:
+:$x \in A^0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(x + y) dx = (x - y) dy}
+Tags: Examples of First Order ODEs, Examples of Homogeneous Differential Equation, First Order ODE: (x + y) dx = (x - y) dy
+
+\begin{theorem}
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|general solution]]:
+:$\arctan \dfrac y x = \ln \sqrt {x^2 + y^2} + C$
+\end{theorem}<|endoftext|>
+\section{First Order ODE/(x^2 - 2 y^2) dx + x y dy = 0}
+Tags: Examples of Homogeneous Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|solution]]:
+:$y^2 = x^2 + C x^4$
+\end{theorem}
+
+\begin{proof}
+$(1)$ can also be rendered:
+:$\dfrac {\mathrm d y} {\mathrm d x} = -\dfrac {x^2 - 2 y^2} {x y}$
+Let:
+:$\map M {x, y} = x^2 - 2 y^2$
+:$\map N {x, y} = x y$
+Put $t x, t y$ for $x, y$:
+{{begin-eqn}}
+{{eqn | l = \map M {t x, t y}
+      | r = \paren {t x}^2 - 2 \paren {t y}^2
+      | c = 
+}}
+{{eqn | r = t^2 \paren {x^2 - 2 y^2}
+      | c = 
+}}
+{{eqn | r = t^2 \, \map M {x, y}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \map N {t x, t y}
+      | r = t x t y
+      | c = 
+}}
+{{eqn | r = t^2 \paren {x y}
+      | c = 
+}}
+{{eqn | r = t^2 \, N \left({x, y}\right)
+      | c = 
+}}
+{{end-eqn}}
+Thus both $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $2$.
+Thus, by definition, $(1)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+By [[Solution to Homogeneous Differential Equation]], its solution is:
+:$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
+where:
+:$\map f {x, y} = -\dfrac {x^2 - 2 y^2} {x y}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {-\frac {1 - 2 z^2} {z} - z} + C_1
+      | c = 
+}}
+{{eqn | r = \int \frac {z \rd z} {z^2 - 1} + C_1
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \, \map \ln {z^2 - 1} + C_1
+      | c = [[Primitive of x over x squared minus a squared|Primitive of $\dfrac x {x^2 - a^2}$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \ln {z^2 - 1}
+      | r = \map \ln {x^2} + \ln C
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z^2 - 1
+      | r = C x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y^2} {x^2} - 1
+      | r = C x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^2
+      | r = x^2 + C x^4
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/x^2 y' - 3 x y - 2 y^2 = 0}
+Tags: Examples of Homogeneous Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|solution]]:
+:$y = C x^2 \paren {x + y}$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = 3 x y + 2 y^2$
+:$\map N {x, y} = x^2$
+Put $t x, t y$ for $x, y$:
+{{begin-eqn}}
+{{eqn | l = \map M {t x, t y}
+      | r = 3 t x t y + 2 \paren {t y}^2
+      | c = 
+}}
+{{eqn | r = t^2 \paren {3 x y + 2 y^2}
+      | c = 
+}}
+{{eqn | r = t^2 \, \map M {x, y}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \map N {t x, t y}
+      | r = \paren {t x}^2
+      | c = 
+}}
+{{eqn | r = t^2 \paren {x^2}
+      | c = 
+}}
+{{eqn | r = t^2 \, \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Thus both $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $2$.
+Thus, by definition, $(1)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+By [[Solution to Homogeneous Differential Equation]], its solution is:
+:$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
+where:
+:$\map f {x, y} = \dfrac {3 x y + 2 y^2} {x^2}$
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {\dfrac {3 z + 2 z^2} {1^2} - z} + C_1
+      | c = 
+}}
+{{eqn | r = \int \frac {\d z} {2 z \paren {1 + z} } + C_1
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \ln \frac z {z + 1} + \ln C_2
+      | c = [[Primitive of Reciprocal of x by a x + b|Primitive of $\dfrac 1 {x \paren {a x + b} }$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \ln x^2
+      | r = \ln \frac z {z + 1} + \ln C_3
+      | c = 
+}}
+{{eqn | r = \ln \frac {C_3 z} {z + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^2
+      | r = \frac {C_3 y / x} {y / x + 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = C x^2 \paren {x + y}
+      | c = where $C := \dfrac 1 {C_3}$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Condensation Points is Monotone}
+Tags: Condensation Points
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $S$.
+Then:
+:$A \subseteq B \implies {A^0} \subseteq B^0$
+where
+:$A^0$ denotes the [[Definition:Set of Condensation Points|set of condensation points]] of $A$
+\end{theorem}
+
+\begin{proof}
+Assume
+:$A \subseteq B$
+Let $x \in A^0$.
+By definition of [[Definition:Set of Condensation Points|set of condensation points]]:
+:$x$ is [[Definition:Condensation Point|condensation point]] of $A$
+By definition of [[Definition:Condensation Point|condensation point]]:
+:$x$ is [[Definition:Limit Point of Set|limit point]] of $A$ such that $\forall U \in \tau: A \cap U$ is [[Definition:Countable Set|uncountable]]
+Thus by [[Limit Point of Subset is Limit Point of Set]]:
+:$x$ is [[Definition:Limit Point of Set|limit point]] of $B$
+Let $U \in \tau$.
+By definition of [[Definition:Condensation Point|condensation point]]:
+:$A \cap U$ is [[Definition:Countable Set|uncountable]]
+By [[Set Intersection Preserves Subsets/Corollary]]:
+:$A \cap U \subseteq B \cap U$
+Thus by [[Subset of Countable Set is Countable]]
+:$B \cap U$ is [[Definition:Countable Set|uncountable]]
+Then by definition:
+:$x$ is [[Definition:Condensation Point|condensation point]] of $B$
+Thus by definition of [[Definition:Set of Condensation Points|set of condensation points]]:
+:$x \in B^0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/x^2 y' = 3 (x^2 + y^2) arctan (y over x) + x y}
+Tags: Examples of Homogeneous Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|solution]]:
+:$y = x \tan C x^3$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = 3 \paren {x^2 + y^2} \arctan \dfrac y x + x y$
+:$\map N {x, y} = x^2$
+Put $t x, t y$ for $x, y$:
+{{begin-eqn}}
+{{eqn | l = \map M {t x, t y}
+      | r = 3 \paren {\paren {t x}^2 + \paren {t y}^2} \arctan \dfrac {t y} {t x} + t x t y
+      | c = 
+}}
+{{eqn | r = t^2 \paren {3 \paren {x^2 + y^2} \arctan \dfrac y x + x y}
+      | c = 
+}}
+{{eqn | r = t^2 \, \map M {x, y}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \map N {t x, t y}
+      | r = \paren {t x}^2
+      | c = 
+}}
+{{eqn | r = t^2 \paren {x^2}
+      | c = 
+}}
+{{eqn | r = t^2 \, \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Thus both $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $2$.
+Thus, by definition, $(1)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+By [[Solution to Homogeneous Differential Equation]], its solution is:
+:$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
+where:
+:$\map f {x, y} = \dfrac {3 \paren {x^2 + y^2} \arctan \dfrac y x + x y} {x^2}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {\dfrac {3 \paren {1^2 + z^2} \arctan \frac z 1 + z} {1^2} - z} + C_1
+      | c = 
+}}
+{{eqn | r = \int \frac {\d z} {3 \paren {1 + z^2} \arctan z} + C_1
+      | c = 
+}}
+{{end-eqn}}
+Substituting $u = \arctan z$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d u} {\d x}
+      | r = \frac 1 {z^2 + 1}
+      | c = [[Derivative of Arctangent Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d z} {\d u}
+      | r = z^2 + 1
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {3 \paren {1 + z^2} \arctan z} + C_1
+      | c = 
+}}
+{{eqn | r = \int \frac {\d u} {3 u} + C_1
+      | c = 
+}}
+{{eqn | r = \frac 1 3 \ln u + C_1
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | ll= \leadsto
+      | l = \ln x^3 + \ln C
+      | r = \ln u
+      | c = putting $\ln C = - C_1$
+}}
+{{eqn | ll= \leadsto
+      | l = C x^3
+      | r = u
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C x^3
+      | r = \arctan \frac y x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = x \tan C x^3
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/x sine (y over x) y' = y sine (y over x) + x}
+Tags: Examples of Homogeneous Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|solution]]:
+:$\cos \dfrac y x + \ln C x = 0$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = y \sin \dfrac y x + x$
+:$\map N {x, y} = x \sin \dfrac y x$
+Put $t x, t y$ for $x, y$:
+{{begin-eqn}}
+{{eqn | l = \map M {t x, t y}
+      | r = t y \sin \dfrac t y t x + t x
+      | c = 
+}}
+{{eqn | r = t \paren {y \sin \dfrac y x + x}
+      | c = 
+}}
+{{eqn | r = t \, \map M {x, y}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \map N {t x, t y}
+      | r = t x \sin \dfrac {t y} {t x}
+      | c = 
+}}
+{{eqn | r = t \paren {x \sin \dfrac y x}
+      | c = 
+}}
+{{eqn | r = t \, \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Thus both $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $1$.
+Thus, by definition, $(1)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+By [[Solution to Homogeneous Differential Equation]], its solution is:
+:$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
+where:
+:$\map f {x, y} = \dfrac {y \sin \dfrac y x + x} {x \sin \dfrac y x}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {\dfrac {z \sin z + 1} {\sin z} - z} + C_1
+      | c = 
+}}
+{{eqn | r = \int \sin z \rd z + C_1
+      | c = 
+}}
+{{eqn | r = -\cos z + C_1
+      | c = [[Primitive of Sine Function|Primitive of $\sin z$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \ln x + \ln C
+      | r = -\cos z
+      | c = putting $\ln C = - C_1$
+}}
+{{eqn | ll= \leadsto
+      | l = \cos \frac y x + \ln C x = 0
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/x y' = y + 2 x exp (- y over x)}
+Tags: Examples of Homogeneous Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|solution]]:
+:$e^{y / x} = \ln x^2 + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = y + 2 x e^{-y/x}$
+:$\map N {x, y} = x$
+Put $t x, t y$ for $x, y$:
+{{begin-eqn}}
+{{eqn | l = \map M {t x, t y}
+      | r = t y + 2 t x e^{-t y / t x}
+      | c = 
+}}
+{{eqn | r = t \paren {y + 2 x e^{-y / x} }
+      | c = 
+}}
+{{eqn | r = t \, \map M {x, y}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \map N {t x, t y}
+      | r = t x
+      | c = 
+}}
+{{eqn | r = t \, \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Thus both $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $1$.
+Thus, by definition, $(1)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+By [[Solution to Homogeneous Differential Equation]], its solution is:
+:$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
+where:
+:$\map f {x, y} = \dfrac {y + 2 x e^{-y / x} } x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {\dfrac {z + 2 e^{-z} } 1 - z} + C_1
+      | c = 
+}}
+{{eqn | r = \int \frac {e^z} 2 \rd z + C_1
+      | c = 
+}}
+{{eqn | r = \frac {e^z} 2 + C_1
+      | c = [[Primitive of Exponential Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \ln x + C
+      | r = e^z
+      | c = putting $C = -\dfrac {C_1} 2$
+}}
+{{eqn | ll= \leadsto
+      | l = e^{y / x}
+      | r = \ln x^2 + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE in form y' = f (a x + b y + c)}
+Tags: First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ODE]]:
+:$\dfrac {\mathrm d y} {\mathrm d x} = f \left({a x + b y + c}\right)$
+can be solved by substituting:
+:$z := a x + b y + c$
+to obtain:
+:$\displaystyle x = \int \frac {\mathrm d z} {b f \left({z}\right) + a}$
+\end{theorem}
+
+\begin{proof}
+We have:
+:$\dfrac {\mathrm d y} {\mathrm d x} = f \left({a x + b y + c}\right)$
+Put:
+: $z := a x + b y + c$
+Then:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = a x + b y + c
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = a + b \dfrac {\mathrm d y} {\mathrm d x}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \dfrac {\mathrm d y} {\mathrm d x}
+      | r = \frac 1 b \left({\frac {\mathrm d z} {\mathrm d x} - a}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac 1 b \left({\frac {\mathrm d z} {\mathrm d x} - a}\right)
+      | r = f \left({z}\right)
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = \frac {\mathrm d z} {\mathrm d x}
+      | r = b f \left({z}\right) + a
+      | c = 
+}}
+{{end-eqn}}
+This can be solved by [[Separation of Variables]]:
+:$\displaystyle x = \int \frac {\mathrm d z} {b f \left({z}\right) + a}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y' = (x + y)^2}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$\dfrac {\d y} {\d x} = \paren {x + y}^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x + y = \map \tan {x + C}$
+\end{theorem}
+
+\begin{proof}
+Make the substitution:
+:$z = x + y$
+Then from [[First Order ODE in form y' = f (a x + b y + c)|First Order ODE in form $y' = f (a x + b y + c)$]] with $a = b = 1$:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \int \frac {\d z} {z^2 + 1}
+      | c = 
+}}
+{{eqn | r = \arctan z + C_1
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | ll= \leadsto
+      | l = y - C_1
+      | r = \map \arctan {x + y}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x + y
+      | r = \map \tan {x + C}
+      | c = letting $C = - C_1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y' = sin^2 (x - y + 1)}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$\dfrac {\d y} {\d x} = \map {\sin^2} {x - y + 1}^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\map \tan {x - y + 1} = x + C$
+\end{theorem}
+
+\begin{proof}
+Make the substitution:
+:$z = x - y + 1$
+Then from [[First Order ODE in form y' = f (a x + b y + c)|First Order ODE in form $y' = f (a x + b y + c)$]] with $a = 1, b = - 1$:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \int \frac {\d z} {- \sin^2 z + 1}
+      | c = 
+}}
+{{eqn | r = \int \frac {\d z} {\cos^2 z}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = \int \sec^2 z \rd z
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \tan z + C_1
+      | c = [[Primitive of Square of Secant Function|Primitive of $\sec^2 z$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \map \tan {x - y + 1}
+      | r = x + C
+      | c = letting $C = - C_1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))}
+Tags: First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ODE]]:
+:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
+such that:
+:$ a e \ne b d$
+can be solved by substituting:
+:$x := z - h$
+:$y := w - k$
+where:
+:$h = \dfrac {c e - b f} {a e - b d}$
+:$k = \dfrac {a f - c d} {a e - b d}$
+to obtain:
+:$\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
+which can be solved by the technique of [[Solution to Homogeneous Differential Equation]].
+\end{theorem}
+
+\begin{proof}
+We have:
+:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
+Make the substitutions:
+:$x := z - h$
+:$y := w - k$
+We have:
+:$\dfrac {\d x} {\d z} = 1$
+:$\dfrac {\d y} {\d w} = 1$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \frac {\d w} {\d z}
+      | r = \map F {\frac {a \paren {z - h} + b \paren {w - k} + c} {d \paren {z - h} + e \paren {w - k} + f} }
+      | c = 
+}}
+{{eqn | r = \map F {\frac {a z + b w - a h - b k + c} {d z + e w - d h - e k + f} }
+      | c = 
+}}
+{{end-eqn}}
+In order to simplify this appropriately, we wish to reduce it to the form:
+:$\dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
+by finding values of $h$ and $k$ such that:
+:$a h + b k = c$
+:$d h + e k = f$
+So:
+{{begin-eqn}}
+{{eqn | l = a h + b k
+      | r = c
+      | c = 
+}}
+{{eqn | l = d h + e k
+      | r = f
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a d h + b d k
+      | r = c d
+      | c = 
+}}
+{{eqn | l = a d h + a e k
+      | r = a f
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k \paren {a e - b d}
+      | r = a f - c d
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = k
+      | r = \frac {a f - c d} {a e - b d}
+      | c = 
+}}
+{{end-eqn}}
+Similarly:
+{{begin-eqn}}
+{{eqn | l = a h + b k
+      | r = c
+      | c = 
+}}
+{{eqn | l = d h + e k
+      | r = f
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = a e h + b e k
+      | r = c e
+      | c = 
+}}
+{{eqn | l = b d h + b e k
+      | r = b f
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = h \paren {a e - b d}
+      | r = c e - b f
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = h
+      | r = \frac {c e - b f} {a e - b d}
+      | c = 
+}}
+{{end-eqn}}
+We note that the above works {{iff}}:
+:$a e - b d \ne 0 \implies a e \ne b d$
+Thus:
+:$(1): \quad \dfrac {\d w} {\d z} = \map F {\dfrac {a z + b w} {d z + e w} }$
+Letting:
+:$\map M {z, w} = a z + b w$
+:$\map N {z, w} = d z + e w$
+we see:
+:$\map M {t z, t w} = a t z + b t w = t \paren {a z + b w} = t \, \map M {z, w}$
+:$\map N {t z, t w} = d t z + e t w = t \paren {d z + e w} = t \, \map N {z, w}$
+Thus, by definition, $(1)$ is [[Definition:Homogeneous Differential Equation|homogeneous]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(x + y + 4) over (x - y - 6)}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x - y - 6}$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\map \arctan {\dfrac {y + 5} {x - 1} } = \ln \sqrt {\paren {x - 1}^2 + \paren {y + 5}^2} + C$
+\end{theorem}
+
+\begin{proof}
+We note that $(1)$ is in the form:
+:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
+where:
+:$a e = -1 \ne b d = 1$
+Hence we can use [[First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))|First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$]].
+Let:
+:$x = z - h$ where $h = 2 / -2 = -1$
+:$y = w - k$ where $k = -10 / -2 = 5$.
+Then:
+:$\dfrac {\d w} {\d z} = \dfrac {z + w} {z - w}$
+From the [[First Order ODE/(x + y) dx = (x - y) dy|solution to $\dfrac {\d y} {\d x} = \dfrac {x + y} {x - y}$]]:
+:$\arctan \dfrac w z = \ln \sqrt {z^2 + w^2} + C$
+We have:
+:$x = z - \paren {-1}$
+:$y = w - 5$
+and so:
+:$z = x - 1$
+:$w = y + 5$
+This gives:
+:$\map \arctan {\dfrac {y + 5} {x - 1} } = \ln \sqrt {\paren {x - 1}^2 + \paren {y + 5}^2} + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(x + y + 4) over (x + y - 6)}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x + y - 6}$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y - x = 5 \, \map \ln {x + y - 1} + C$
+\end{theorem}
+
+\begin{proof}
+We note that $(1)$ is in the form:
+:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
+where:
+:$a e = b d = 1$
+Hence we can use [[First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d|First Order ODE in form $y' = \map F {\dfrac{a x + b y + c} {d x + e y + f} }$ where $a e = b d$]].
+Let:
+:$z = x + y$
+to obtain:
+:$\dfrac {\d z} {\d x} = b \, \map F {\dfrac {a z + a c} {d z + f} } + a$
+where:
+:$a = 1$
+:$c = 4$
+:$d = 1$
+:$f = -6$
+which gives:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d z} {\d x}
+      | r = \dfrac {z + 4} {z - 6} + 1
+      | c = 
+}}
+{{eqn | r = \frac {z + 4 + z - 6} {z - 6}
+      | c = 
+}}
+{{eqn | r = \frac {2 \paren {z - 1} } {z - 6}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int \rd x
+      | r = \int \frac {z - 6} {2 \paren {z - 1} } \rd z
+      | c = 
+}}
+{{end-eqn}}
+Substitute $v = z - 1$ which gives:
+:$\dfrac {\d z} {\d v} = 1$
+and thence:
+:$v = x + y - 1$:
+{{begin-eqn}}
+{{eqn | l = \int \rd x
+      | r = \frac 1 2 \int \frac {v - 5} v \rd v
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x
+      | r = \frac v 2 - \frac 5 2 \ln v + C_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 x
+      | r = v - 5 \ln v + C_2
+      | c = 
+}}
+{{end-eqn}}
+Substituting $x + y - 1$ for $v$:
+{{begin-eqn}}
+{{eqn | l = 2 x
+      | r = x + y - 1 - 5 \, \map \ln {x + y - 1} + C_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x - y
+      | r = -5 \, \map \ln {x + y - 1} + C_3
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y - x
+      | r = 5 \, \map \ln {x + y - 1} + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f)) where a e = b d/Example}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+=== [[First Order ODE/(x + y + 4) over (x + y - 6)|$\dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x + y - 6}$]] ===
+{{:First Order ODE/(x + y + 4) over (x + y - 6)}}
+\end{theorem}<|endoftext|>
+\section{First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))/Example}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+=== [[First Order ODE/(x + y + 4) over (x - y - 6)|$\dfrac {\d y} {\d x} = \dfrac {x + y + 4} {x - y - 6}$]] ===
+{{:First Order ODE/(x + y + 4) over (x - y - 6)}}
+\end{theorem}<|endoftext|>
+\section{First Order ODE/exp y dx + (x exp y + 2 y) dy = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$x e^y + y^2 = C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = e^y$
+:$\map N {x, y} = x e^y + 2 y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = e^y
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = e^y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int e^y \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = x e^y + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {x e^y + 2 y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = x e^y + y^2 + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = x e^y + y^2$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$x e^y + y^2 = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(x + (2 over y)) dy + y dx = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$x y + 2 \ln y = C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = y$
+:$\map N {x, y} = x + \dfrac 2 y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = 1
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int y \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = x y + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {x + \dfrac 2 y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = x y + 2 \ln y + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = x y + 2 \ln y$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$x y + 2 \ln y = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(y - x^3) dx + (x + y^3) dy = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$4 x y - x^4 + y^4 = C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = y - x^3$
+:$\map N {x, y} = x + y^3$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = 1
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {y - x^3} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = x y - \dfrac {x^4} 4 + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \left({x + y^3}\right) \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = x y + \dfrac {y^4} 4 + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = x y - \dfrac {x^4} 4 + \dfrac {y^4} 4$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$x y - \dfrac {x^4} 4 + \dfrac {y^4} 4 = C_1$
+which can be simplified by multiplying through by $4$ and setting $C = 4 C_1$:
+:$4 x y - x^4 + y^4 = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(y + y cosine x y) dx + (x + x cosine x y) dy = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$x y + \sin x y = C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = y + y \cos x y$
+:$\map N {x, y} = x + x \cos x y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = 1 + \cos x y  - x \sin x y
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = 1 + \cos x y  - x \sin x y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {y + y \cos x y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = x y + \sin x y + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {x + x \cos x y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = x y + \sin x y + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = x y + \sin x y$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$x y + \sin x y = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(sine x sine y - x e^y) dy = (e^y + cosine x cosine y) dx}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$\sin x \cos y + x e^y = C$
+\end{theorem}
+
+\begin{proof}
+First express $(1)$ in the form:
+:$(2): \quad -\paren {e^y + \cos x \cos y} + \paren {\sin x \sin y - x e^y} \dfrac {\d y} {\d x}$
+Let:
+:$\map M {x, y} = -\paren {e^y + \cos x \cos y}$
+:$\map N {x, y} = \sin x \sin y - x e^y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = -e^y + \cos x \sin y
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = -e^y + \cos x \sin y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = -\int \paren {e^y + \cos x \cos y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = -x e^y - \sin x \cos y + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {\sin x \sin y - x e^y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = -\sin x \cos y - x e^y + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = -\sin x \cos y - x e^y$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$, after simplification, is:
+:$\sin x \cos y + x e^y = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/-1 over y sine x over y dx + x over y^2 sine x over y dy}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$\dfrac x y = C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = -\dfrac 1 y \sin \dfrac x y$
+:$\map N {x, y} = \dfrac x {y^2} \sin \dfrac x y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\partial M} {\partial y}
+      | r = \frac 1 {y^2} \sin \frac x y + \paren {-\frac 1 y} \paren {-\frac 1 {y^2} } x \cos \frac x y
+      | c = 
+}}
+{{eqn | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^3} \cos \frac x y
+      | c = 
+}}
+{{eqn | l = \frac {\partial N} {\partial x}
+      | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^2} \frac 1 y \cos \frac x y
+      | c = 
+}}
+{{eqn | r = \frac 1 {y^2} \sin \frac x y + \frac x {y^3} \cos \frac x y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r =\map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {-\dfrac 1 y \sin \dfrac x y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = -\dfrac 1 y y \cos \dfrac x y + \map g y
+      | c = 
+}}
+{{eqn | r = -\cos \dfrac x y + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {\dfrac x {y^2} \sin \dfrac x y} \rd y + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Set $z = \dfrac x y$:
+{{begin-eqn}}
+{{eqn | l = z
+      | r = \frac x y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x}
+      | r = -\frac x {y^2}
+      | c = 
+}}
+{{end-eqn}}
+Thus by [[Integration by Substitution]]:
+{{begin-eqn}}
+{{eqn | l = \int \frac x {y^2} \sin \frac x y \rd y
+      | r = \int \frac x {y^2} \frac {-y^2} x \sin z \rd z
+      | c = 
+}}
+{{eqn | r = -\int \sin z \rd z
+      | c = 
+}}
+{{eqn | r = \cos z
+      | c = 
+}}
+{{eqn | r = \cos \frac x y
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = -\cos \dfrac x y$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$, after simplification, is:
+{{begin-eqn}}
+{{eqn | l = \cos \dfrac x y
+      | r = C_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \arccos {\cos \dfrac x y}
+      | r = \arccos C_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac x y
+      | r = C
+      | c = where $C = \arccos C_1$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(2 x y^3 + y cosine x) dx + (3 x^2 y^2 + sine x) dy}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$x^2 y^3 + y \sin x = C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = 2 x y^3 + y \cos x$
+:$\map N {x, y} = 3 x^2 y^2 + \sin x$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\partial M} {\partial y}
+      | r = 2 x \cdot 3 y^2 + \cos x
+      | c = 
+}}
+{{eqn | r = 6 x y^2 + \cos x
+      | c = 
+}}
+{{eqn | l = \frac {\partial N} {\partial x}
+      | r = 3 x^2 \cdot 2 y^2 + \cos x
+      | c = 
+}}
+{{eqn | r = 6 x y^2 + \cos x
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {2 x y^3 + y \cos x} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = x^2 y^3 + y \sin x + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {3 x^2 y^2 + \sin x} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = x^2 y^3 + y \sin x + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = x^2 y^3 + y \sin x$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$x^2 y^3 + y \sin x = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/dx = (y over (1 - x^2 y^2)) dx + (x over (1 - x^2 y^2)) dy}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$\map \ln {\dfrac {1 + x y} {1 - x y} } - 2 x = C$
+\end{theorem}
+
+\begin{proof}
+First express $(1)$ in the form:
+:$(2): \quad \paren {\dfrac y {1 - x^2 y^2} - 1} + \paren {\dfrac x {1 - x^2 y^2} } \dfrac {\d y} {\d x}$
+Let:
+:$\map M {x, y} = \dfrac y {1 - x^2 y^2} - 1$
+:$\map N {x, y} = \dfrac x {1 - x^2 y^2}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\partial M} {\partial y}
+      | r = \frac 1 {1 - x^2 y^2} + \frac {2 x^2 y^2} {\paren {1 - x^2 y^2}^2}
+      | c = 
+}}
+{{eqn | r = \frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2}
+      | c = 
+}}
+{{eqn | l = \frac {\partial N} {\partial x}
+      | r = \frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2}
+      | c = similarly
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {\dfrac y {1 - x^2 y^2} - 1} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \dfrac {y \rd x} {1 - x^2 y^2} - \int \rd x + \map g y
+      | c = 
+}}
+{{end-eqn}}
+Substitute $z = x y$ to obtain:
+:$\dfrac {\d z} {\d x} = y$
+This gives:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \map \ln {\frac {1 + z} {1 - z} } - \int \rd x + \map g y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map M {x, y} \rd x + \map g y
+      | r = \frac 1 2 \map \ln {\frac {1 + x y} {1 - x y} } - x + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {\frac {1 + x^2 y^2} {\paren {1 - x^2 y^2}^2} } \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \map \ln {\frac {1 + x y} {1 - x y} } + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = \dfrac 1 2 \map \ln {\dfrac {1 + x y} {1 - x y} } - x$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$, after simplification, is:
+:$\map \ln {\dfrac {1 + x y} {1 - x y} } - 2 x = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Heart Curve}
+Tags: Analytic Geometry
+
+\begin{theorem}
+The following equations define [[Definition:Curve|curves]] that are more or less heart shaped:
+\end{theorem}<|endoftext|>
+\section{Heart Curve/Formulation 1}
+Tags: Heart Curves
+
+\begin{theorem}
+:$r = 1 - \sin \theta$
+:[[File:HeartCurve1.png|500px]]
+\end{theorem}<|endoftext|>
+\section{Heart Curve/Formulation 2}
+Tags: Heart Curves
+
+\begin{theorem}
+:$\left({x^2 + y^2 - 1}\right)^3 = x^2 y^3$
+:[[File:HeartCurve2.png|500px]]
+\end{theorem}<|endoftext|>
+\section{Heart Curve/Formulation 3}
+Tags: Heart Curves
+
+\begin{theorem}
+:$\begin{cases}
+x = & \map \sin t \map \cos t \ln \size t \\
+y = & \size t^{0.3} \sqrt {\cos t}
+\end{cases}$
+where $t \in \hointr {-1} 0 \cup \hointl 0 1$.
+:[[File:HeartCurve3.png|500px]]
+\end{theorem}<|endoftext|>
+\section{Limit Point of Subset is Limit Point of Set}
+Tags: Limit Points
+
+\begin{theorem}
+Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
+Let $A, B$ be [[Definition:Subset|subset]] of $S$ such that
+:$A \subseteq B$
+Let $x$ be a [[Definition:Element|point]] of $S$.
+Then:
+:if $x$ is [[Definition:Limit Point of Set|limit point]] of $A$, then $x$ is [[Definition:Limit Point of Set|limit point]] of $B$.
+\end{theorem}
+
+\begin{proof}
+Assume $x$ is [[Definition:Limit Point of Set|limit point]] of $A$.
+By definition of [[Definition:Limit Point of Set|limit point]] it suffices to prove
+:$\forall U \in \tau: x \in U \implies B \cap \paren {U \setminus \set x} \ne \O$
+Let $U \in \tau$ such that
+:$x \in U$
+By definition of [[Definition:Limit Point of Set|limit point]]:
+:$A \cap \paren {U \setminus \set x} \ne \O$
+By [[Set Intersection Preserves Subsets/Corollary]]:
+:$A \cap \paren {U \setminus \set x} \subseteq B \cap \paren {U \setminus \set x}$
+Thus:
+:$B \cap \paren {U \setminus \set x} \ne \O$
+{{qed}}
+[[Category:Limit Points]]
+9wavooswmawzbhvl8fgyr4757e2ghdf
+\end{proof}<|endoftext|>
+\section{Heart Curve/Formulation 4}
+Tags: Heart Curves
+
+\begin{theorem}
+:$\left({y - \dfrac {2 \left({\left\lvert{x}\right\rvert + x^2 - 6}\right)} {3 \left({\left\lvert{x}\right\rvert + x^2 + 2}\right)} }\right)^2 + x^2 = 36$
+:[[File:HeartCurve4.png|500px]]
+\end{theorem}<|endoftext|>
+\section{Heart Curve/Formulation 5}
+Tags: Heart Curves
+
+\begin{theorem}
+:$r = \dfrac {\sin t \sqrt {\size {\cos t} } } {\sin t + \frac 7 5} - 2 \sin t + 2$
+:[[File:HeartCurve5.png|500px]]
+\end{theorem}<|endoftext|>
+\section{Heart Curve/Formulation 6}
+Tags: Heart Curves
+
+\begin{theorem}
+:$\begin{cases}
+x = & 16 \sin^3 t \\
+y = & 13 \cos t - 5 \cos 2 t - 3 \cos 3 t - \cos 4 t
+\end{cases}$
+:[[File:HeartCurve6.png|500px]]
+\end{theorem}<|endoftext|>
+\section{Existence of Infinitely Many Integrating Factors}
+Tags: Differential Equations
+
+\begin{theorem}
+Let the [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
+be such that $M$ and $N$ are [[Definition:Real Function|real functions]] of two variables which are ''not'' [[Definition:Homogeneous Real Function|homogeneous functions]] of the same [[Definition:Degree of Homogeneous Real Function|degree]].
+Suppose $(1)$ has an [[Definition:Integrating Factor|integrating factor]].
+Then $(1)$ has an [[Definition:Infinite|infinite number]] of [[Definition:Integrating Factor|integrating factors]]
+\end{theorem}
+
+\begin{proof}
+Let $\map F f$ be any function of $f$ which is an [[Definition:Integrating Factor|integrating factor]] of $(1)$.
+Then:
+:$\ds \mu \map F f \paren {\map M {x, y} \rd x + \map N {x, y} \rd y} = \map F f \rd f = \map \d {\int \map F f \rd f}$
+so $\mu \map F f$ is also an [[Definition:Integrating Factor|integrating factor]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integrating Factor for First Order ODE/Preliminary Work}
+Tags: Integrating Factors
+
+\begin{theorem}
+Let the [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
+be non-[[Definition:Homogeneous Differential Equation|homogeneous]] and not [[Definition:Exact Differential Equation|exact]].
+Let $\map \mu x$ be an [[Definition:Integrating Factor|integrating factor]] for $(1)$.
+Let:
+:$\map P {x, y} := \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
+Then:
+:$\dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M  \dfrac {\partial \mu} {\partial y} }$
+\end{theorem}
+
+\begin{proof}
+Let us for ease of manipulation express $(1)$ in the form of [[Definition:Differential|differentials]]:
+:$(2): \quad \map M {x, y} \rd x + \map N {x, y} \rd y = 0$
+Let $\mu$ be an [[Definition:Integrating Factor|integrating factor]] for $(2)$.
+Then, by definition:
+:$\mu \, \map M {x, y} \rd x + \mu \, \map N {x, y} \rd y = 0$
+is an [[Definition:Exact Differential Equation|exact differential equation]].
+By [[Solution to Exact Differential Equation]]:
+:$\dfrac {\map \partial {\mu M} } {\partial y} = \dfrac {\map \partial {\mu N} } {\partial x}$
+Evaluating this, using the [[Product Rule]]:
+:$\mu \dfrac {\partial M} {\partial y} + M \dfrac {\partial \mu} {\partial y} = \mu \dfrac {\partial N} {\partial x} + N \dfrac {\partial \mu} {\partial x}$
+which leads us to:
+:$\dfrac 1 \mu \paren {N \dfrac {\partial \mu} {\partial x} - M  \dfrac {\partial \mu} {\partial y} } = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
+Let us use $\map P {x, y}$ for $\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$.
+Then:
+:$(3): \quad \dfrac 1 \mu = \dfrac {\map P {x, y} } {N \dfrac {\partial \mu} {\partial x} - M \dfrac {\partial \mu} {\partial y}}$
+\end{proof}<|endoftext|>
+\section{Integrating Factor for First Order ODE/Function of One Variable}
+Tags: Integrating Factors
+
+\begin{theorem}
+Suppose that:
+:$g \left({x}\right) = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {N \left({x, y}\right)}$
+is a [[Definition:Real Function|function]] of $x$ only.
+Then:
+:$\mu \left({x}\right) = e^{\int g \left({x}\right) \mathrm d x}$
+is an [[Definition:Integrating Factor|integrating factor]] for $(1)$.
+Similarly, suppose that:
+:$h \left({y}\right) = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {M \left({x, y}\right)}$
+is a [[Definition:Real Function|function]] of $y$ only.
+Then:
+:$\mu \left({y}\right) = e^{\int -h \left({y}\right) \mathrm d y}$
+is an [[Definition:Integrating Factor|integrating factor]] for $(1)$.
+\end{theorem}
+
+\begin{proof}
+=== [[Integrating Factor for First Order ODE/Preliminary Work|Preliminary Work]] ===
+{{:Integrating Factor for First Order ODE/Preliminary Work}}
+=== Proof for Function of $x$ or $y$ only ===
+Suppose that $\mu$ is a function of $x$ only.
+Then:
+:$\dfrac {\partial \mu} {\partial x} = \dfrac {d \mu} {d x}, \dfrac {\partial \mu} {\partial y} = 0$
+which, when substituting in $(3)$, leads us to:
+:$\dfrac 1 \mu \dfrac {\mathrm d \mu} {\mathrm d x} = \dfrac {P \left({x, y}\right)} {N \left({x, y}\right)} = g \left({x}\right)$
+where $g \left({x}\right)$ is the function of $x$ that we posited.
+Similarly, if $\mu$ is a function of $y$ only, we find that
+:$\dfrac 1 \mu \dfrac {\mathrm d \mu} {\mathrm d y} = \dfrac {P \left({x, y}\right)} {-M \left({x, y}\right)} = h \left({y}\right)$
+where $h \left({y}\right)$ is the function of $y$ that we posited.
+=== [[Integrating Factor for First Order ODE/Conclusion|Conclusion of Proof]] ===
+{{:Integrating Factor for First Order ODE/Conclusion}}
+\end{proof}<|endoftext|>
+\section{Integrating Factor for First Order ODE/Function of Sum of Variables}
+Tags: Integrating Factors
+
+\begin{theorem}
+Suppose that:
+:$g \left({z}\right) = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x} } {N \left({x, y}\right) - M \left({x, y}\right)}$
+is a [[Definition:Real Function|function]] of $z$, where $z = x + y$.
+Then:
+:$\mu \left({x + y}\right) = \mu \left({z}\right) = e^{\int g \left({z}\right) \mathrm d z}$
+is an [[Definition:Integrating Factor|integrating factor]] for $(1)$.
+\end{theorem}
+
+\begin{proof}
+=== [[Integrating Factor for First Order ODE/Preliminary Work|Preliminary Work]] ===
+{{:Integrating Factor for First Order ODE/Preliminary Work}}
+=== Proof for Function of $x + y$ ===
+Suppose that $\mu$ is a function of $z = x + y$.
+Then:
+:$\dfrac {\partial z} {\partial x} = 1 = \dfrac {\partial z} {\partial y}$
+Thus:
+:$\dfrac {\partial \mu} {\partial x} = \dfrac {d \mu} {d z} \dfrac {\partial z} {\partial x} = \dfrac {\mathrm d \mu} {\mathrm d z} = \dfrac {\mathrm d \mu} {\mathrm d z} \dfrac {\partial z} {\partial y} = \dfrac {\partial \mu} {\partial y}$
+which, when substituting in $(3)$, leads us to:
+:$\dfrac 1 \mu \dfrac {\mathrm d \mu} {\mathrm d z} = \dfrac {P \left({x, y}\right)} {N \left({x, y}\right) - M \left({x, y}\right)} = g \left({z}\right)$
+where $g \left({z}\right)$ is the function of $z$ that we posited.
+=== [[Integrating Factor for First Order ODE/Conclusion|Conclusion of Proof]] ===
+{{:Integrating Factor for First Order ODE/Conclusion}}
+\end{proof}<|endoftext|>
+\section{Integrating Factor for First Order ODE/Function of Product of Variables}
+Tags: Integrating Factors
+
+\begin{theorem}
+:$\map g z = \dfrac {\dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}} {N y - M x}$
+is a [[Definition:Real Function|function]] of $z$, where $z = x y$.
+Then:
+:$\map \mu {x y} = \map \mu z = e^{\int \map g z \d z}$
+is an [[Definition:Integrating Factor|integrating factor]] for $(1)$.
+\end{theorem}
+
+\begin{proof}
+=== [[Integrating Factor for First Order ODE/Preliminary Work|Preliminary Work]] ===
+{{:Integrating Factor for First Order ODE/Preliminary Work}}
+=== Proof for Function of x y ===
+Suppose that $\mu$ is a function of $z = x y$.
+Then:
+:$\dfrac {\partial z} {\partial x} = y$
+:$\dfrac {\partial z} {\partial y} = x$
+Thus:
+:$\dfrac {\partial \mu} {\partial x} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial x} = y \dfrac {\d \mu} {\d z}$
+:$\dfrac {\partial \mu} {\partial y} = \dfrac {\d \mu} {\d z} \dfrac {\partial z} {\partial y} = x \dfrac {\d \mu} {\d z}$
+which, when substituting in $(3)$, leads us to:
+:$\dfrac 1 \mu \dfrac {\d \mu} {\d z} = \dfrac {\map P {x, y} } {N y - M x} = \map g z$
+where $\map g z$ is the function of $z$ that we posited.
+=== [[Integrating Factor for First Order ODE/Conclusion|Conclusion of Proof]] ===
+{{:Integrating Factor for First Order ODE/Conclusion}}
+\end{proof}<|endoftext|>
+\section{Set of Condensation Points of Union is Union of Sets of Condensation Points}
+Tags: Condensation Points
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $S$.
+Then:
+:$\left({A \cup B}\right)^0 = A^0 \cup B^0$
+\end{theorem}
+
+\begin{proof}
+=== Set of Condensation Points of Union Subset Union of Sets of Condensation Points ===
+Let $x \in \left({A \cup B}\right)^0$.
+By definition of [[Definition:Set of Condensation Points|set of condensation points]]:
+:$x$ is [[Definition:Condensation Point|condensation point]] of $A \cup B$
+By [[Set of Condensation Points of Union is Union of Sets of Condensation Points/Lemma|Lemma]]:
+:$x$ is [[Definition:Condensation Point|condensation point]] of $A$ or $x$ is [[Definition:Condensation Point|condensation point]] of $B$
+By definition of [[Definition:Set of Condensation Points|set of condensation points]]:
+:$x \in A^0$ or $x \in B^0$
+Thus by definition of [[Definition:Set Union|union]]:
+:$x \in A^0 \cup B^0$
+{{qed|lemma}}
+=== Union of Sets of Condensation Points Subset Set of Condensation Points of Union ===
+By [[Set is Subset of Union]]:
+:$A \subseteq A \cup B \land B \subseteq A \cup B$
+By [[Set of Condensation Points is Monotone]]:
+:$A^0 \subseteq \left({A \cup B}\right)^0 \land B^0 \subseteq \left({A \cup B}\right)^0$
+Thus by [[Union is Smallest Superset]]:
+:$A^0 \cup B^0 \subseteq \left({A \cup B}\right)^0$
+{{qed|lemma}}
+Thus by definition of [[Definition:Set Equality|set equality]]:
+:$\left({A \cup B}\right)^0 = A^0 \cup B^0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Condensation Points of Union is Union of Sets of Condensation Points/Lemma}
+Tags: Condensation Points
+
+\begin{theorem}
+Let $x$ be a [[Definition:Element|point]] of $S$.
+Then:
+:if $x$ is [[Definition:Condensation Point|condensation point]] of $A \cup B$,
+:then $x$ is [[Definition:Condensation Point|condensation point]] of $A$ or $x$ is [[Definition:Condensation Point|condensation point]] of $B$.
+\end{theorem}
+
+\begin{proof}
+Assume $x$ is [[Definition:Condensation Point|condensation point]] of $A \cup B$.
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$x$ is not [[Definition:Condensation Point|condensation point]] of $A$ and $x$ is not [[Definition:Condensation Point|condensation point]] of $B$.
+By definition of [[Definition:Condensation Point|condensation point]]:
+:$\exists U_1 \in \tau: x \in U_1 \land A \cap U_1$ is [[Definition:Countable Set|countable]]
+By definition of [[Definition:Condensation Point|condensation point]]:
+:$\exists U_2 \in \tau: x \in U_2 \land B \cap U_2$ is [[Definition:Countable Set|countable]]
+By definition of [[Definition:Set Intersection|intersection]]:
+:$x \in U_1 \cap U_2$
+By definition of [[Definition:Topological Space|topological space]]:
+:$U_1 \cap U_2 \in \tau$
+By [[Intersection is Subset]]:
+:$A \cap U_1 \cap U_2 \subseteq A \cap U_1 \land B \cap U_1 \cap U_2 \subseteq B \cap U_2$
+By [[Set Union Preserves Subsets]]:
+:$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) \subseteq \left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$
+By [[Countable Union of Countable Sets is Countable]]:
+:$\left({A \cap U_1}\right) \cup \left({B \cap U_2}\right)$ is [[Definition:Countable Set|countable]]
+Then by [[Subset of Countable Set is Countable]]:
+:$(1): \quad \left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right)$ is [[Definition:Countable Set|countable]]
+By [[Intersection Distributes over Union]]:
+:$\left({A \cap U_1 \cap U_2}\right) \cup \left({B \cap U_1 \cap U_2}\right) = \left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$
+By definition of [[Definition:Condensation Point|condensation point]]:
+:$\left({A \cup B}\right) \cap \left({U_1 \cap U_2}\right)$ is [[Definition:Countable Set|uncountable]]
+This contradicts $(1)$.
+Thus the result fallows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integrating Factor for First Order ODE/Conclusion}
+Tags: Integrating Factors
+
+\begin{theorem}
+Let the [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \map M {x, y} + \map N {x, y} \dfrac {\d y} {\d x} = 0$
+be non-[[Definition:Homogeneous Differential Equation|homogeneous]] and not [[Definition:Exact Differential Equation|exact]].
+Let $\map \mu {x, y}$be an [[Definition:Integrating Factor|integrating factor]] for $(1)$.
+If one of these is the case:
+:$\mu$ is a function of $x$ only
+:$\mu$ is a function of $y$ only
+:$\mu$ is a function of $x + y$
+:$\mu$ is a function of $x y$
+then:
+:$\mu = e^{\int \map f w \rd w}$
+where $w$ depends on the nature of $\mu$.
+\end{theorem}
+
+\begin{proof}
+We have an equation of the form:
+:$\dfrac 1 \mu \dfrac {\d \mu} {\d w} = \map f w$
+which is what you get when you apply the [[Chain Rule for Derivatives]] and [[Derivative of Logarithm Function]] to:
+:$\dfrac {\map \d {\ln \mu} } {\d w} = \map f w$
+Thus:
+:$\displaystyle \ln \mu = \int \map f w \rd w$
+and so:
+:$\mu = e^{\int \map f w \rd w}$
+Hence the results as stated.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y dx + (x^2 y - x) dy = 0}
+Tags: Examples of First Order ODEs, First Order ODE: y dx + (x^2 y - x) dy = 0
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad y \rd x + \paren {x^2 y - x} \rd y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\dfrac {y^2} 2 - \dfrac y x = C$
+\end{theorem}<|endoftext|>
+\section{Set of Condensation Points of Countable Set is Empty}
+Tags: Condensation Points
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[Definition:Topological Space|topological space]].
+Let $A$ be a [[Definition:Subset|subset]] of $S$.
+Then:
+:if $A$ is [[Definition:Countable Set|countable]],
+:then $A^0 = \varnothing$.
+\end{theorem}
+
+\begin{proof}
+Assume
+:$A$ is [[Definition:Countable Set|countable]].
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that
+:$A^0 \ne \varnothing$
+By definition of [[Definition:Empty Set|empty set]]:
+:$\exists x: x \in A^0$
+Then by definition of [[Definition:Set of Condensation Points|set of condensation points]]:
+:$x$ is a [[Definition:Condensation Point|condensation point]] of $A$.
+This contradicts [[Set of Condensation Points of Countable Set is Empty/Lemma|Lemma]].
+Thus the result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Set of Condensation Points of Countable Set is Empty/Lemma}
+Tags: Condensation Points
+
+\begin{theorem}
+:if $A$ is [[Definition:Countable Set|countable]],
+:then there exists no [[Definition:Element|point]] $x$ of $S$ such that $x$ is a [[Definition:Condensation Point|condensation point]] of $A$.
+\end{theorem}
+
+\begin{proof}
+Assume
+:$A$ is [[Definition:Countable Set|countable]].
+Aiming for a [[Definition:Contradiction|contradiction]] suppose that there exists a [[Definition:Element|point]] $x$ of $S$ such that
+: $x$ is a [[Definition:Condensation Point|condensation point]] of $A$
+By definition of [[Definition:Topological Space|topological space]]:
+:$S \in \tau$
+Then by definition of [[Definition:Condensation Point|condensation point]]:
+:$A \cap S$ is [[Definition:Countable Set|uncountable]]
+By [[Intersection with Subset is Subset]]:
+:$A \cap S = A$
+$A$ is [[Definition:Countable Set|countable]] contradicts $A$ is [[Definition:Countable Set|uncountable]].
+Thus the result follows by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(3 x^2 - y^2) dy - 2 x y dx = 0}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {3 x^2 - y^2} \rd y - 2 x y \rd x = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
+\end{theorem}
+
+\begin{proof}
+We note that $(1)$ is in the form:
+:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
+but that $(1)$ is not [[Definition:Exact Differential Equation|exact]].
+So, let:
+:$\map M {x, y} = -2 x y$
+:$\map N {x, y} = 3 x^2 - y^2$
+Let:
+:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map P {x, y}
+      | r = -2 x - 6 x
+      | c = 
+}}
+{{eqn | r = -8 x
+      | c = 
+}}
+{{end-eqn}}
+It can be observed that:
+{{begin-eqn}}
+{{eqn | l = \frac {\map P {x, y} } {\map M {x, y} }
+      | r = \frac {-8 x} {-2 x y}
+      | c = 
+}}
+{{eqn | r = \frac 4 y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\map P {x, y}} {\map M {x, y} }$ is a function of $y$ only.
+So [[Integrating Factor for First Order ODE/Function of One Variable|Integrating Factor for First Order ODE: Function of One Variable]] can be used:
+:$\map \mu y = e^{\int -\map h y \rd y}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int -\map h y \rd y
+      | r = \int -\paren {4 / y} \rd y
+      | c = 
+}}
+{{eqn | r = - 4 \ln y
+      | c = 
+}}
+{{eqn | r = \map \ln {y^{-4} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int -\map h y \rd y}
+      | r = \frac 1 {y^4}
+      | c = 
+}}
+{{end-eqn}}
+Thus an [[Definition:Integrating Factor|integrating factor]] for $(1)$ has been found:
+:$\mu = \dfrac 1 {y^4}$
+which yields, when multiplying it throughout $(1)$:
+:$\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$
+which is now [[Definition:Exact Differential Equation|exact]].
+By [[First Order ODE/(3 x^2 over y^4 - 1 over y^2) dy - 2 x over y^3 dx = 0|First Order ODE: $\paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$]], its solution is:
+:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(x y - 1) dx + (x^2 - x y) dy = 0}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {x y - 1} \rd x + \paren {x^2 - x y} \rd y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x y - \ln x - \dfrac {y^2} 2 + C$
+\end{theorem}
+
+\begin{proof}
+We note that $(1)$ is in the form:
+:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
+but that $(1)$ is not [[Definition:Exact Differential Equation|exact]].
+So, let:
+:$\map M {x, y} = x y - 1$
+:$\map N {x, y} = x^2 - x y = x \paren {x - y}$
+Let:
+:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map P {x, y}
+      | r = x - \paren {2 x - y}
+      | c = 
+}}
+{{eqn | r = -\paren {x - y}
+      | c = 
+}}
+{{end-eqn}}
+It can be observed that:
+{{begin-eqn}}
+{{eqn | l = \frac {\map P {x, y} } {\map N {x, y} }
+      | r = \frac {-\paren {x - y} } {x \paren {x - y} }
+      | c = 
+}}
+{{eqn | r = -\frac 1 x
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\map P {x, y} } {\map N {x, y}}$ is a function of $x$ only.
+So [[Integrating Factor for First Order ODE/Function of One Variable|Integrating Factor for First Order ODE: Function of One Variable]] can be used:
+:$\map \mu y = e^{\int \map g x \rd x}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int \map g x \rd x
+      | r = \int -\paren {1 / x} \rd x
+      | c = 
+}}
+{{eqn | r = -\ln x
+      | c = 
+}}
+{{eqn | r = \map \ln {x^{-1} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int \map g x \rd x}
+      | r = \frac 1 x
+      | c = 
+}}
+{{end-eqn}}
+Thus an [[Definition:Integrating Factor|integrating factor]] for $(1)$ has been found:
+:$\mu = \dfrac 1 x$
+which yields, when multiplying it throughout $(1)$:
+:$\paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$
+which is now [[Definition:Exact Differential Equation|exact]].
+By [[First Order ODE/(y - 1 over x) dx + (x - y) dy = 0|First Order ODE: $\paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$]], its solution is:
+:$x y - \ln x - \dfrac {y^2} 2 + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y dx + x dy + 3 x^3 y^4 dy}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad y \rd x + x \rd y + 3 x^3 y^4 \rd y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
+\end{theorem}
+
+\begin{proof}
+We note that $(1)$ is in the form:
+:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
+but that $(1)$ is not [[Definition:Exact Differential Equation|exact]].
+So, let:
+:$\map M {x, y} = y$
+:$\map N {x, y} = 3 x^3 y^4 + x = x \paren {3 x^2 y^4 + 1}$
+Let:
+:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map P {x, y}
+      | r = 1 - \paren {9 x^2 y^4 + 1}
+      | c = 
+}}
+{{eqn | r = -9 x^2 y^4
+      | c = 
+}}
+{{end-eqn}}
+It can be observed that:
+{{begin-eqn}}
+{{eqn | l = \frac {\map P {x, y} } {\map N {x, y} y - \map M {x, y} x}
+      | r = \frac {-9 x^2 y^4} {\paren {3 x^3 y^4 + x} y - y x}
+      | c = 
+}}
+{{eqn | r = -\frac {-9 x^2 y^4} {3 x^3 y^5}
+      | c = 
+}}
+{{eqn | r = -\frac 3 {x y}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\map P {x, y} } {\map M {x, y} }$ is a function of $x y$.
+So [[Integrating Factor for First Order ODE/Function of Product of Variables|Integrating Factor for First Order ODE: Function of Product of Variables]] can be used.
+Let $z = x y$.
+Then:
+:$\map \mu {x y} = \map \mu z = e^{\int \map g z \rd z}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int \map g z \rd z
+      | r = \int -\frac 3 z \rd z
+      | c = 
+}}
+{{eqn | r = -3 \ln z
+      | c = 
+}}
+{{eqn | r = \map \ln {z^{-3} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int \map g z \rd z}
+      | r = \frac 1 {z^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {x^3 y^3}
+      | c = 
+}}
+{{end-eqn}}
+Thus an [[Definition:Integrating Factor|integrating factor]] for $(1)$ has been found:
+:$\mu = \dfrac 1 {x^3 y^3}$
+which yields, when multiplying it throughout $(1)$:
+:$\dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$
+which is now [[Definition:Exact Differential Equation|exact]].
+By [[First Order ODE/1 over x^3 y^2 dx + (1 over x^2 y^3 + 3 y) dy = 0|First Order ODE: $\dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$]], its solution is:
+:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Curved Mirror producing Parallel Rays is Paraboloid}
+Tags: Optics
+
+\begin{theorem}
+Let $M$ be a [[Definition:Curve|curved]] [[Definition:Mirror|mirror]] embedded in a [[Definition:Real Cartesian Space|real cartesian $3$- space]].
+Let there be a source of [[Definition:Light|light]] at the [[Definition:Origin|origin]].
+Let $M$ reflect the [[Definition:Light|light]] in a beam [[Definition:Parallel Lines|parallel]] to the [[Definition:X-Axis|$x$-axis]].
+Then $M$ is the [[Definition:Solid of Revolution|solid of revolution]] produced by rotating about the [[Definition:X-Axis|$x$-axis]] the [[Definition:Parabola|parabola]] whose equation is:
+:$y^2 = 2 c x + c^2$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Mirror|mirror]] will have the shape of a [[Definition:Surface of Revolution|surface of revolution]] generated by revolving a [[Definition:Curve|curve]] $APB$ in the [[Definition:Cartesian Plane|cartesian plane]] around the [[Definition:X-Axis|$x$-axis]].
+Let $P = \left({x, y}\right)$ be an arbitrary [[Definition:Point|point]] on $APB$.
+:[[File:ParabolicMirror.png|500px]]
+From the [[Law of Reflection]]:
+:$\alpha = \beta$
+By the geometry of the situation:
+:$\phi = \beta$
+:$\theta = \alpha + \phi = 2 \beta$
+By definition of [[Definition:Tangent of Angle|tangent]]:
+:$\tan \theta = \dfrac y x$
+and so:
+{{begin-eqn}}
+{{eqn | l = \tan \theta
+      | r = \tan 2 \beta
+      | c = 
+}}
+{{eqn | r = \dfrac {2 \tan \beta} {1 - \tan^2 \beta}
+      | c = [[Double Angle Formula for Tangent]]
+}}
+{{eqn | ll= \implies
+      | l = \frac y x
+      | r = \dfrac {2 \dfrac {\mathrm d y} {\mathrm d x} } {1 - \left({\dfrac {\mathrm d y} {\mathrm d x} }\right)^2}
+      | c = [[Double Angle Formula for Tangent]]
+}}
+{{end-eqn}}
+Using the [[Quadratic Formula]]:
+:$\dfrac {\mathrm d y} {\mathrm d x} = \dfrac {-x \pm \sqrt {x^2 + y^2} \, \mathrm d x} y$
+which can be expressed as:
+:$x \, \mathrm d x + y  \, \mathrm d y = \pm \sqrt {x^2 + y^2} \, \mathrm d x$
+Using [[Differential of Sum of Squares]]:
+:$\pm \dfrac {\mathrm d \left({x^2 + y^2}\right)} {2 \sqrt {x^2 + y^2} } = \mathrm d x$
+and so:
+:$\pm \sqrt {x^2 + y^2} = x + c$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/x dy - y dx = (1 + y^2) dy}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x \rd y - y \rd x = \paren {1 + y^2} \rd y$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\dfrac x y = \dfrac 1 y - y + C$
+\end{theorem}
+
+\begin{proof}
+Rearranging, we have:
+:$\dfrac {y \rd x - x \rd y} {y^2} = -\paren {\dfrac 1 {y^2} + 1} \rd y$
+From the [[Quotient Rule for Derivatives]]:
+:$\map \d {\dfrac x y} = \dfrac {y \rd x - x \rd y} {y^2}$
+from which:
+:$\map \d {\dfrac x y} = -\paren {\dfrac 1 {y^2} + 1} \rd y$
+Hence the result:
+:$\dfrac x y = \dfrac 1 y - y + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Quotient Rule for Differentials/Formulation 1}
+Tags: Differentials
+
+\begin{theorem}
+:$\map \d {\dfrac y x} = \dfrac {x \rd y - y \rd x} {x^2}$
+\end{theorem}<|endoftext|>
+\section{Quotient Rule for Differentials/Formulation 2}
+Tags: Differentials
+
+\begin{theorem}
+:$\mathrm d \left({\dfrac x y}\right) = \dfrac {y \, \mathrm d x - x \, \mathrm d y} {y^2}$
+\end{theorem}<|endoftext|>
+\section{Product Rule for Differentials}
+Tags: Differentials
+
+\begin{theorem}
+:$\map \d {x y} = x \rd y + y \rd x$
+\end{theorem}<|endoftext|>
+\section{Differential of Sum of Squares}
+Tags: Differentials
+
+\begin{theorem}
+:$\map \d {x^2 + y^2} = 2 \paren {x \map \rd x + y \map \rd y}$
+\end{theorem}<|endoftext|>
+\section{Differential of Arctangent of Quotient}
+Tags: Differentials
+
+\begin{theorem}
+:$\mathrm d \left({\arctan \dfrac x y}\right) = \dfrac{y \, \mathrm d x - x \, \mathrm d y} {x^2 + y^2}$
+\end{theorem}<|endoftext|>
+\section{Differential of Logarithm of Quotient}
+Tags: Differentials
+
+\begin{theorem}
+:$\map \d {\ln \dfrac x y} = \dfrac {y \rd x - x \rd y} {x y}$
+\end{theorem}<|endoftext|>
+\section{First Order ODE/y dx + (x^2 y - x) dy = 0/Proof 1}
+Tags: First Order ODE: y dx + (x^2 y - x) dy = 0
+
+\begin{theorem}
+{{:First Order ODE/y dx + (x^2 y - x) dy = 0}}
+This can also be presented in the form:
+:$\dfrac {\d y} {\d x} + \dfrac y {x^2 y - x}$
+\end{theorem}
+
+\begin{proof}
+We note that $(1)$ is in the form:
+:$\map M {x, y} \d x + \map N {x, y} \d y = 0$
+but that $(1)$ is not [[Definition:Exact Differential Equation|exact]].
+So, let:
+:$\map M {x, y} = y$
+:$\map N {x, y} = x^2 y - x$
+Let:
+:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map P {x, y}
+      | r = 1 - \paren {2 x y - 1}
+      | c = 
+}}
+{{eqn | r = 2 \paren {1 - x y}
+      | c = 
+}}
+{{end-eqn}}
+It can be observed that:
+{{begin-eqn}}
+{{eqn | l = \frac {\map P {x, y} } {\map N {x, y} }
+      | r = \frac {2 \paren {1 - x y} } {x^2 y - x}
+      | c = 
+}}
+{{eqn | r = \frac {-2 \paren {x y - 1} } {x \paren {x y - 1} }
+      | c = 
+}}
+{{eqn | r = -\frac 2 x
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\map P {x, y} } {\map N {x, y} }$ is a function of $x$ only.
+So [[Integrating Factor for First Order ODE/Function of One Variable|Integrating Factor for First Order ODE: Function of One Variable]] can be used:
+:$\map \mu x = e^{\int \map g x \rd x}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int \map g x \rd x
+      | r = \int - \paren {2 / x} \rd x
+      | c = 
+}}
+{{eqn | r = - 2 \ln x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int \map g x \rd x}
+      | r = \frac 1 {x^2}
+      | c = 
+}}
+{{end-eqn}}
+Thus an [[Definition:Integrating Factor|integrating factor]] for $(1)$ has been found:
+:$\mu = \dfrac 1 {x^2}$
+which yields, when multiplying it throughout $(1)$:
+:$\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$
+which is now [[Definition:Exact Differential Equation|exact]].
+By [[First Order ODE/(y over x^2) dx + (y - 1 over x) dy = 0|First Order ODE: $\dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$]], its solution is:
+:$\dfrac {y^2} 2 - \dfrac y x = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y dx - x dy = x y^3 dy}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad y rd x - x \rd y = x y^3 \rd y$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\ln \dfrac x y = \dfrac {y^3} 3 + C$
+\end{theorem}
+
+\begin{proof}
+Rearranging, we have:
+:$\dfrac {y \rd x - x \rd y} {x y} = y^2 \rd y$
+From [[Differential of Logarithm of Quotient]]:
+:$\map \d {\ln \dfrac y x} = \dfrac {y \rd x - x \rd y} {x y}$
+from which:
+:$\map \d {\ln \dfrac x y} = y^2 \rd y$
+Hence the result:
+:$\ln \dfrac x y = \dfrac {y^3} 3 + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/x dy = (x^5 + x^3 y^2 + y) dx}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x \rd y = \paren {x^5 + x^3 y^2 + y} \rd x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\arctan \dfrac x y = -\dfrac {x^4} 4 + C$
+\end{theorem}
+
+\begin{proof}
+Rearranging, we have:
+:$y \rd x - x \rd y = -\paren {x^2 + y^2} x^3 \rd x$
+from which:
+:$\dfrac {y \rd x - x \rd y} {x^2 + y^2} = - x^3 \rd x$
+From [[Differential of Arctangent of Quotient]]:
+:$\map \d {\arctan \dfrac x y} = \dfrac {y \rd x - x \rd y} {x^2 + y^2}$
+from which $(1)$ evolves into:
+:$\map \d {\arctan \dfrac x y} = -x^3 \rd x$
+Hence the result:
+:$\arctan \dfrac x y = -\dfrac {x^4} 4 + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(x + y) dx = (x - y) dy/Proof 1}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \paren {x + y} \rd x = \paren {x - y} \rd y$
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|solution]]:
+:$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} =  x + y$
+:$\map N {x, y} =  x - y$
+We have that:
+:$\map M {t x, t y} =  t x + t y = t \paren {x + y} = t \map M {x, y}$
+:$\map N {t x, t y} =  t x - t y = t \paren {x - y} = t \map N {x, y}$
+Thus both $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $1$.
+Thus by definition $(1)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+{{begin-eqn}}
+{{eqn | l = \frac {\d y} {\d x}
+      | r = \frac {x + y} {x - y}
+      | c = 
+}}
+{{eqn | r = \frac {1 + y / x} {1 - y / x}
+      | c = dividing [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $x$
+}}
+{{eqn | r = \frac {1 + z} {1 - z}
+      | c = substituting $z$ for $y / x$
+}}
+{{end-eqn}}
+By [[Solution to Homogeneous Differential Equation]]:
+:$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
+where:
+:$\map f {1, z} = \dfrac {1 + z} {1 - z}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {\dfrac {1 + z} {1 - z} - z}
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {1 - z} \rd z} {1 + z - z \paren {1 - z} }
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {1 - z} \rd z} {1 + z^2}
+      | c = 
+}}
+{{eqn | r = \int \frac {\d z} {1 + z^2} - \int \frac {z \rd z} {1 + z^2}
+      | c = 
+}}
+{{eqn | r = \arctan z - \int \frac {z \rd z} {1 + z^2} + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | r = \arctan z - \frac 1 2 \map \ln {1 + z^2} + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{end-eqn}}
+Substituting $y / x$ for $z$ reveals the solution:
+:$\arctan \dfrac y x = \ln \sqrt{x^2 + y^2} + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/x dy = (y + x^2 + 9 y^2) dx}
+Tags: Examples of First Order ODEs, First Order ODE: x dy = (y + x^2 + 9 y^2) dx
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x \rd y = \paren {y + x^2 + 9 y^2} \rd x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\map \arctan {\dfrac {3 y} x} = 3 x + C$
+\end{theorem}<|endoftext|>
+\section{Linear First Order ODE/y' + (y over x) = 3 x}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$\dfrac {\d y} {\d x} + \dfrac y x = 3 x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x y = x^3 + C$
+or:
+:$y = x^2 + \dfrac C x$
+\end{theorem}
+
+\begin{proof}
+This is a special case of:
+:[[Linear First Order ODE/y' + (y over x) = k x^n|Linear First Order ODE: $\dfrac {\d y} {\d x} + \dfrac y x = k x^n$]]
+where $k = 3$ and $n = 1$, yielding:
+:$y = x^2 + \dfrac C x$
+Multiplying through by $x$ reveals:
+:$x y = x^3 + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/x y' - 3 y = x^4}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad x \dfrac {\d y} {\d x} - 3y = x^4$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = x^4 + \dfrac C {x^3}$
+\end{theorem}
+
+\begin{proof}
+Rearranging $(1)$:
+:$(2): \quad \dfrac {\d y} {\d x} + \paren {-\dfrac 3 x} y = x^3$
+$(2)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where:
+:$\map P x = -\dfrac 3 x$
+:$\map Q x = x^3$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int -\frac 3 x \rd x
+      | c = 
+}}
+{{eqn | r = -3 \ln x
+      | c = 
+}}
+{{eqn | r = \ln x^{-3}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = e^{\ln x^{-3} }
+      | c = 
+}}
+{{eqn | r = x^{-3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {x^3}
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\map {\dfrac \d {\d x} } {\dfrac y {x^3} } = 1$
+and the [[Definition:General Solution|general solution]] is:
+:$\dfrac y {x^3} = x + C$
+or:
+:$y = x^4 + \dfrac C {x^3}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/y' + y = 1 over (1 + exp 2 x)}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad y' + y = \dfrac 1 {1 + e^{2 x} }$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = e^{-x} \map \arctan {e^x} + C e^{-x}$
+\end{theorem}
+
+\begin{proof}
+$(1)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where $\map P x = 1$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int \rd x
+      | c = 
+}}
+{{eqn | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = e^x
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\map {\dfrac \d {\d x} } {e^x y} = \dfrac {e^x} {1 + e^{2 x} }$
+and the [[Definition:General Solution|general solution]] becomes:
+:$\displaystyle y {e^x} = \int \frac {e^x} {1 + e^{2 x} } \rd x$
+The integral on the {{RHS}} can be solved by substituting:
+:$u = e^x \implies \d u = e^x \rd x$
+and so:
+{{begin-eqn}}
+{{eqn | l = e^x y
+      | r = \int \frac {1 \rd u} {1 + u^2}
+      | c = 
+}}
+{{eqn | r = \map \arctan u + C
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | r = \map \arctan {e^x} + C
+      | c = 
+}}
+{{end-eqn}}
+or:
+:$y = e^{-x} \map \arctan {e^x} + C e^{-x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/(1 + x^2) dy + 2 x y dx = cotangent x dx}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad \paren {1 + x^2} \rd y + 2 x y \rd x = \cot x \rd x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \dfrac {\map \ln {\sin x} } {1 + x^2} + \dfrac C {1 + x^2}$
+\end{theorem}
+
+\begin{proof}
+$(1)$ can be written as:
+:$(2): \quad \paren {1 + x^2} \dfrac {\rd y} {\rd x} + 2 x y = \cot x$
+We notice straight away that:
+:$\dfrac {\rd} {\rd x} \paren {1 + x^2} = 2 x$
+and so:
+:$\dfrac {\rd} {\rd x} \paren {1 + x^2} y = \cot x$
+Thus the [[Definition:General Solution of Differential Equation|general solution]] becomes:
+{{begin-eqn}}
+{{eqn | l = \paren {1 + x^2} y
+      | r = \int \cot x \rd x
+      | c = 
+}}
+{{eqn | r = \map \ln {\sin x} + C
+      | c = 
+}}
+{{end-eqn}}
+or:
+:$y = \dfrac {\map \ln {\sin x} } {1 + x^2} + \dfrac C {1 + x^2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/y' + y = 2 x exp -x + x^2}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad y' + y = 2 x e^{-x} + x^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = x^2 e^{-x} + x^2 - 2 x + 2 + C e^{-x}$
+\end{theorem}
+
+\begin{proof}
+$(1)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where:
+:$\map P x = 1$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int 1 \rd x
+      | c = 
+}}
+{{eqn | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = e^x
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\dfrac \d {\d x} e^x y = 2 x + x^2 e^x$
+and the [[Definition:General Solution|general solution]] is:
+:$e^x y = x^2 + e^x \paren {x^2 - 2 x + 2} + C$
+or:
+:$y = x^2 e^{-x} + x^2 - 2 x + 2 + C e^{-x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/y' + y cot x = 2 x cosec x}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad y' + y \cot x = 2 x \csc x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = x^2 \csc x + C \csc x$
+\end{theorem}
+
+\begin{proof}
+$(1)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where:
+:$\map P x = \cot x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int \cot x \rd x
+      | c = 
+}}
+{{eqn | r = \map \ln {\sin x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = \sin x
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\dfrac {\d} {\d x} y \sin x = 2 x$
+and the [[Definition:General Solution|general solution]] is:
+:$y \sin x = x^2 + C$
+or:
+:$y = x^2 \csc x + C \csc x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/(2 y - x^3) dx = x dy}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad \paren {2 y - x^3} \rd x = x \rd y$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = -x^3 + C x^2$
+\end{theorem}
+
+\begin{proof}
+Rearranging $(1)$:
+:$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = - x^2$
+$(2)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where $\map P x = -\dfrac 2 x$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int -\dfrac 2 x \rd x
+      | c = 
+}}
+{{eqn | r = -2 \ln x
+      | c = 
+}}
+{{eqn | r = \map \ln {\frac 1 {x^2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = \frac 1 {x^2}
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\map {\dfrac \d {\d x} } {\dfrac y {x^2} } = -1$
+and the [[Definition:General Solution|general solution]] is:
+:$\dfrac y {x^2} = -x + C$
+or:
+:$y = -x^3 + C x^2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Solution to Bernoulli's Equation}
+Tags: Bernoulli's Equation
+
+\begin{theorem}
+'''[[Definition:Bernoulli's Equation|Bernoulli's equation]]''':
+:$(1): \quad \dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
+where:
+:$n \ne 0, n \ne 1$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$\displaystyle \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
+where:
+:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
+\end{theorem}
+
+\begin{proof}
+Make the substitution:
+:$z = y^{1 - n}$
+in $(1)$.
+Then we have:
+{{begin-eqn}}
+{{eqn | l = \frac {\d z} {\d y}
+      | r = \paren {1 - n} y^{-n}
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d y} \frac {\d y} {\d x} + \map P x y \paren {1 - n} y^{-n}
+      | r = \map Q x y^n \paren {1 - n} y^{-n}
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x} + \paren {1 - n} \map P x y^{1 - n}
+      | r = \paren {1 - n} \map Q x
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d z} {\d x} + \paren {1 - n} \map P x z
+      | r = \paren {1 - n} \map Q x
+      | c =
+}}
+{{end-eqn}}
+This is now a [[Definition:Linear First Order Ordinary Differential Equation|linear first order ordinary differential equation]] in $z$.
+It has an [[Definition:Integrating Factor|integrating factor]]:
+{{begin-eqn}}
+{{eqn | l = \map \mu x
+      | r = e^{\int \paren {1 - n} \map P x \rd x}
+}}
+{{eqn | r = e^{\paren {1 - n} \int \map P x \rd x}
+      | c =
+}}
+{{end-eqn}}
+and this can be used to obtain:
+:$\displaystyle \map \mu x z = \paren {1 - n} \int \map Q x \map \mu x \rd x + C$
+Substituting $z = y^{1 - n} = \dfrac 1 {y^{n - 1} }$ finishes the proof.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernoulli's Equation/x y' + y = x^4 y^3}
+Tags: Examples of Bernoulli's Equation
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x y' + y = x^4 y^3$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$\dfrac 1 {y^2} = - x^4 + C x^2$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be rearranged as:
+:$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = x^3 y^3$
+It can be seen that $(2)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
+where:
+:$\map P x = \dfrac 1 x$
+:$\map Q x = x^3$
+:$n = 3$
+and so is an example of [[Definition:Bernoulli's Equation|Bernoulli's equation]].
+By [[Solution to Bernoulli's Equation]] it has the [[Definition:General Solution|general solution]]:
+:$(3): \quad \displaystyle \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$
+where:
+:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
+Thus $\map \mu x$ is evaluated:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - n} \int \map P x \rd x
+      | r = \paren {1 - 3} \int \dfrac 1 x \rd x
+      | c = 
+}}
+{{eqn | r = -2 \ln x
+      | c = 
+}}
+{{eqn | r = \map \ln {\frac 1 {x^2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \mu x
+      | r = e^{\map \ln {\frac 1 {x^2} } }
+      | c = 
+}}
+{{eqn | r = \frac 1 {x^2}
+      | c = 
+}}
+{{end-eqn}}
+and so substituting into $(3)$:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {x^2} \frac 1 {y^2}
+      | r = \paren {1 - 3} \int x^3 \frac 1 {x^2} \rd x
+      | c = 
+}}
+{{eqn | r = -2 \int x \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {x^2 y^2}
+      | r = -2 \dfrac {x^2} 2 + C
+      | c = 
+}}
+{{end-eqn}}
+Hence the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$ is:
+:$\dfrac 1 {y^2} = - x^4 + C x^2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernoulli's Equation/x y^2 y' + y^3 = x cosine x}
+Tags: Examples of Bernoulli's Equation
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x y^2 y' + y^3 = x \cos x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y^3 = 3 \sin x + \dfrac {9 \cos x} x - \dfrac {18 \sin x} {x^2} - \dfrac {18 \cos x} {x^3} + \dfrac C {x^3}$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be rearranged as:
+:$(2): \quad \dfrac {\d y} {\d x} + \dfrac 1 x y = \dfrac {\cos x} {y^2}$
+It can be seen that $(2)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
+where:
+:$\map P x = \dfrac 1 x$
+:$\map Q x = \cos x$
+:$n = -2$
+and so is an example of [[Definition:Bernoulli's Equation|Bernoulli's equation]].
+By [[Solution to Bernoulli's Equation]] it has the [[Definition:General Solution|general solution]]:
+:$(3): \quad \displaystyle \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$
+where:
+:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
+Thus $\map \mu x$ is evaluated:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - n} \int \map P x \rd x
+      | r = \paren {1 - \paren {-2} } \int \dfrac 1 x \rd x
+      | c = 
+}}
+{{eqn | r = 3 \ln x
+      | c = 
+}}
+{{eqn | r = \ln x^3
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \mu x
+      | r = e^{\ln x^3}
+      | c = 
+}}
+{{eqn | r = x^3
+      | c = 
+}}
+{{end-eqn}}
+and so substituting into $(3)$:
+{{begin-eqn}}
+{{eqn | l = x^3 \frac 1 {y^{-3} }
+      | r = \paren {1 - \paren {-2} } \int x^3 \cos x \rd x
+      | c = 
+}}
+{{eqn | r = 3 \int x^3 \cos x \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x^3 y^3
+      | r = 3 \paren {\paren {3 x^2 - 6} \cos x + \paren {x^3 - 6 x} \sin x} + C
+      | c = [[Primitive of x cubed by Cosine of a x|Primitive of $x^3 \cos a x$]]
+}}
+{{end-eqn}}
+Hence the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$ is:
+:$y^3 = 3 \sin x + \dfrac {9 \cos x} x - \dfrac {18 \sin x} {x^2} - \dfrac {18 \cos x} {x^3} + \dfrac C {x^3}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(exp y - 2 x y) y' = y^2}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {e^y - 2 x y} y' = y^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x y^2 = e^y + C$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be rearranged as:
+:$\dfrac {\d y} {\d x} = \dfrac {y^2} {e^y - 2 x y}$
+Hence:
+:$(2): \quad \dfrac {\d x} {\d y} + \dfrac 2 y x = \dfrac {e^y} {y^2}$
+It can be seen that $(2)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form:
+:$\dfrac {\d x} {\d y} + \map P y x = \map Q y$
+where:
+:$\map P y = \dfrac 2 y$
+:$\map Q y = \dfrac {e^y} {y^2}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P y \rd y
+      | r = \int \dfrac 2 y \rd y
+      | c = 
+}}
+{{eqn | r = 2 \ln y
+      | c = 
+}}
+{{eqn | r = \ln y^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd y}
+      | r = y^2
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(2)$ can be rewritten as:
+:$\map {\dfrac \d {\d y} } {x y^2} = e^y$
+and the [[Definition:General Solution|general solution]] is:
+:$x y^2 = e^y + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y - x y' = y' y^2 exp y}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad y - x y' = y' y^2 e^y$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x y^2 = e^y + C$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be rearranged as:
+:$\dfrac {\d y} {\d x} = \dfrac y {y^2 e^y + x}$
+Hence:
+:$(2): \quad \dfrac {\d x} {\d y} - \dfrac 1 y x = y e^y$
+It can be seen that $(2)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form:
+:$\dfrac {\d x} {\d y} + \map P y x = \map Q y$
+where:
+:$\map P y = -\dfrac 1 y$
+:$\map Q y = y e^y$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P y \rd y
+      | r = \int -\dfrac 1 y \rd y
+      | c = 
+}}
+{{eqn | r = -\ln y
+      | c = 
+}}
+{{eqn | r = \ln y^{-1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd y}
+      | r = \dfrac 1 y
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(2)$ can be rewritten as:
+:$\map {\dfrac {\d} {\d y} } {\dfrac x y} = e^y$
+and the [[Definition:General Solution|general solution]] is:
+:$\dfrac x y = e^y + C$
+or:
+:$x = y e^y + C y$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topology Defined by Neighborhood System}
+Tags: Topology
+
+\begin{theorem}
+Let $S$ be a [[Definition:Set|set]].
+Let $\family {\NN_x}_{x \mathop \in S}$ be an [[Definition:Indexed Family|indexed family]] where $\NN_x$ is [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $S$.
+Assume that
+:$(\text N 1): \quad \forall x \in S, U \in \NN_x: x \in U$
+:$(\text N 2): \quad \forall x \in S, U \in \NN_x, y \in U:\exists V \in \NN_y: V \subseteq U$
+:$(\text N 3): \quad \forall x \in S, U_1, U_2 \in \NN_x: \exists U \in \NN_x: U \subseteq U_1 \cap U_2$
+Then $T = \struct {S, \tau}$ is a [[Definition:Topological Space|topological space]] where:
+:$\tau = \displaystyle \set {\bigcup \GG: \GG \subseteq \bigcup_{x \mathop \in S} \NN_x}$
+Moreover, $\family {\NN_x}_{x \mathop \in S}$ is a [[Definition:Neighborhood System|neighborhood system]] of $T$.
+\end{theorem}
+
+\begin{proof}
+Define:
+:$\BB := \ds \bigcup_{x \mathop \in S} \NN_x$
+According to [[Topology Defined by Basis]] it should be proved that
+:$(\text B 1): \quad \forall A_1, A_2 \in \BB: \forall x \in A_1 \cap A_2: \exists A \in \BB: x \in A \subseteq A_1 \cap A_2$
+:$(\text B 2): \quad \forall x \in S: \exists A \in \BB: x \in A$
+Ad $(\text B 1)$:
+Let $A_1, A_2 \in \BB$.
+Let $x \in A_1 \cap A_2$.
+By definition of [[Definition:Set Intersection|intersection]]:
+:$x \in A_1 \land x \in A_2$
+By definition of [[Definition:Union of Family|union]]:
+:$\exists x_1 \in S: A_1 \in \NN_{x_1}$
+and
+:$\exists x_2 \in S: A_2 \in \NN_{x_2}$
+By $(\text N 2)$:
+:$\exists V_1 \in \NN_x: V_1 \subseteq A_1$
+and
+:$\exists V_2 \in \NN_x: V_2 \subseteq A_2$
+By $(\text N 3)$:
+:$\exists U \in \NN_x: U \subseteq V_1 \cap V_2$
+By definition of [[Definition:Union of Family|union]]:
+:$U \in \BB$
+By [[Set Intersection Preserves Subsets]]:
+:$V_1 \cap V_2 \subseteq A_1 \cap A_2$
+Thus by [[Subset Relation is Transitive]]:
+:$U \subseteq A_1 \cap A_2$
+Ad $(\text B 2)$:
+Let $x \in S$.
+By definition of [[Definition:Empty Set|empty set]]:
+:$\exists U: U \in \NN_x$
+By $(\text N 1)$:
+:$x \in U$
+By definition of [[Definition:Union of Family|union]]:
+:$U \in \BB$
+Thus:
+:$\exists A \in \BB: x \in A$
+By definition of $\tau$ and $\BB$:
+:$\tau = \set {\bigcup \GG: \GG \subseteq \BB}$
+Thus by [[Topology Defined by Basis]]:
+:$T = \struct {S, \tau}$ is a [[Definition:Topological Space|topological space]]
+It remains to prove that
+:$\family {\NN_x}_{x \mathop \in S}$ is a [[Definition:Neighborhood System|neighborhood system]]
+Let $x \in S$.
+Let $U \in \tau$.
+By definition of $\tau$:
+:$\exists \GG \subseteq \BB: U = \bigcup \GG$
+Let $x \in U$.
+By definition of [[Definition:Union of Set of Sets|union]]:
+:$\exists V \in \GG: x \in V$
+By definition of [[Definition:Subset|subset]]:
+:$V \in \BB$
+By definition of [[Definition:Union of Family|union]]:
+:$\exists y \in S: V \in \NN_y$
+By $(\text N 2)$:
+:$\exists W \in \NN_x: W \subseteq V$
+By [[Set is Subset of Union/Set of Sets]]:
+:$V \subseteq U$
+Thus by [[Subset Relation is Transitive]]:
+:$\exists W \in \NN_x: W \subseteq U$
+Thus by definition:
+:$\NN_x$ is a [[Definition:Local Basis|local basis]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Orthogonal Trajectories/x + C exp -x}
+Tags: Orthogonal Trajectories, Exponential Function
+
+\begin{theorem}
+Consider the [[Definition:One-Parameter Family of Curves|one-parameter family of curves]]:
+:$(1): \quad y = x + C e^{-x}$
+Its [[Definition:Orthogonal Trajectories|family of orthogonal trajectories]] is given by the equation:
+:$x = y - 2 + C e^{-y}$
+:[[File:XplusCExpMinusXOrthogonalTrajectories.png|600px]]
+\end{theorem}
+
+\begin{proof}
+We use the technique of [[Definition:Formation of Ordinary Differential Equation by Elimination|formation of ordinary differential equation by elimination]].
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT|Differentiation}} $x$ gives:
+:$\dfrac {\d y} {\d x} = 1 - C e^{-x}$
+Eliminating $C$:
+{{begin-eqn}}
+{{eqn | l = C
+      | o = =
+      | r = \frac {y - x} {e^{-x} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = 1 - \frac {y - x} {e^{-x} } e^{-x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x} + y
+      | r = x + 1
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Orthogonal Trajectories of One-Parameter Family of Curves]], the [[Definition:Orthogonal Trajectories|family of orthogonal trajectories]] is given by:
+:$-\dfrac {\d y} {\d x} = - x = 1 - y$
+The [[Definition:Integrating Factor|integrating factor]] is $e^y$, giving:
+:$\displaystyle e^y x = \int y e^y - e^y \rd y$
+Using [[Primitive of x by Exponential of a x|Primitive of $x e^{a x}$]]:
+:$\displaystyle \int y e^y \rd y = y e^y - e^y$
+Thus we get:
+:$e^y x = y e^y - e^y  - e^y + C$
+which gives us:
+:$x = y - 2 + C e^{-y}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Differential Equation of Family of Linear Combination of Functions is Linear}
+Tags: Linear First Order ODEs
+
+\begin{theorem}
+Consider the [[Definition:One-Parameter Family of Curves|one-parameter family of curves]]:
+:$(1): \quad y = C \map f x + \map g x$
+The [[Definition:Differential Equation|differential equation]] that describes $(1)$ is [[Definition:Linear First Order ODE|linear and of first order]].
+\end{theorem}
+
+\begin{proof}
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT}} $x$ gives:
+:$(2): \quad \dfrac {\d y} {\d x} = C \map {f'} x + \map {g'} x$
+Rearranging $(1)$, we have:
+:$C = \dfrac {y - \map g x} {\map f x}$
+Substituting for $C$ in $(2)$:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d x}
+      | r = \dfrac {y - \map g x} {\map f x} \map {f'} x + \map {g'} x
+      | c = 
+}}
+{{eqn | r = \dfrac {\map {f'} x} {\map f x} y - \dfrac {\map g x \map {f'} x} {\map f x} + \map g x
+      | c = 
+}}
+{{end-eqn}}
+which leaves:
+:$\dfrac {\d y} {\d x} - \dfrac {\map {f'} x} {\map f x} y = \map g x \paren {1 - \dfrac {\map {f'} x} {\map f x} }$
+which is [[Definition:Linear First Order ODE|linear and of first order]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Solution of Second Order Differential Equation with Missing Dependent Variable}
+Tags: Second Order ODEs
+
+\begin{theorem}
+Let $\map f {x, y', y''} = 0$ be a [[Definition:Second Order Ordinary Differential Equation|second order ordinary differential equation]] in which the [[Definition:Dependent Variable|dependent variable]] $y$ is not explicitly present.
+Then $f$ can be reduced to a [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]], whose [[Definition:General Solution|solution]] can be determined.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Second Order Ordinary Differential Equation|second order ordinary differential equation]]:
+:$(1): \quad \map f {x, y', y''} = 0$
+Let a new [[Definition:Dependent Variable|dependent variable]] $p$ be introduced:
+:$y' = p$
+:$y'' = \dfrac {\d p} {\d x}$
+Then $(1)$ can be transformed into:
+:$(2): \quad \map f {x, p, \dfrac {\d p} {\d x} } = 0$
+which is a [[Definition:First Order ODE|first order ODE]].
+If $(2)$ has a [[Definition:General Solution|solution]] which can readily be found, it will be expressible in the form:
+:$(3): \quad \map g {x, p}$
+which can then be expressed in the form:
+:$\map g {x, \dfrac {\d y} {\d x} } = 0$
+which is likewise subject to the techniques of solution of a [[Definition:First Order ODE|first order ODE]].
+Hence such a [[Definition:Second Order ODE|second order ODE]] is reduced to the problem of solving two [[Definition:First Order ODE|first order ODEs]] in succession.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/x y'' - y' = 3 x^2}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x y'' - y' = 3 x^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = x^3 + \dfrac {C_1 x^2} 2 + C^2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]].
+Substitute $p$ for $y'$ in $(1)$:
+:$x \dfrac {\d p} {\d x} - p = 3 x^2$
+and divide through by $x$:
+:$\dfrac {\d p} {\d x} - \dfrac p x = 3 x$
+From:
+:[[Linear First Order ODE/y' - (y over x) = 3 x|Linear First Order ODE: $y' - \dfrac y x = 3 x$]]
+its [[Definition:General Solution to Differential Equation|solution]] is:
+:$p = 3 x^2 + C_1 x$
+Substituting back for $p$:
+:$\dfrac {\d y} {\d x} = 3 x^2 + C_1 x$
+which is [[Separation of Variables|separable]], leading to:
+:$y = x^3 + \dfrac {C_1 x^2} 2 + C^2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Niemytzki Plane is Topology}
+Tags: Niemytzki Plane
+
+\begin{theorem}
+[[Definition:Niemytzki Plane|Niemytzki plane]] is a [[Definition:Topological Space|topological space]].
+\end{theorem}
+
+\begin{proof}
+By definition $T = \struct {S, \tau}$ is the [[Definition:Niemytzki Plane|Niemytzki plane]] {{iff}}:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = S
+      | r = \set {\tuple {x, y}: y \ge 0}
+}}
+{{eqn | n = 2
+      | l = \map \BB {x, y}
+      | r = \set {\map {B_r} {x, y} \cap S: r > 0}
+      | c = if $x, y \in \R, y > 0$
+}}
+{{eqn | n = 3
+      | l = \map \BB {x, 0}
+      | r = \set {\map {B_r} {x, r} \cup \set {\tuple {x, 0} }: r > 0}
+      | c = if $x \in \R$
+}}
+{{eqn | n = 4
+      | l = \tau
+      | r = \set {\bigcup \GG: \GG \subseteq \bigcup_{\tuple {x, y} \in S} \map \BB {x, y} }
+}}
+{{end-eqn}}
+where $\map {B_r} {x, y}$ denotes the [[Definition:Open Ball|open $r$-ball]] of $\tuple {x, y}$ in the [[Definition:Real Number Plane with Euclidean Metric|real number plane $\R^2$ with the usual (Euclidean) metric]].
+According to [[Topology Defined by Neighborhood System]] it should be proved:
+:$(\text N 0): \quad \forall \tuple {x, y} \in S: \map \BB {x, y}$ is [[Definition:Empty Set|non-empty]] [[Definition:Set of Sets|set]] of [[Definition:Subset|subsets]] of $S$
+:$(\text N 1): \quad \forall \tuple {x, y} \in S, U \in \map \BB {x, y}: \tuple {x, y} \in U$
+:$(\text N 2): \quad \forall \tuple {x, z} \in S, U \in \map \BB {x, z}, \tuple {y, s} \in U: \exists V \in \map \BB {y, s}: V \subseteq U$
+:$(\text N 3): \quad \forall \tuple {x, y} \in S, U_1, U_2 \in \map \BB {x, y}: \exists U \in \map \BB {x, y}: U \subseteq U_1 \cap U_2$
+Ad $(\text N 0)$:
+Let $\tuple {x, y} \in S$.
+In both cases: $y > 0$ and $y = 0$, by $(2)$ and $(3)$:
+:$\map \BB {x, y}$ is [[Definition:Non-Empty Set|non-empty]]
+Let $U \in \map \BB {x, y}$.
+In a case when $y > 0$:
+:$\exists r > 0: U = \map {B_r} {x, y} \cap S$
+By [[Intersection is Subset]]:
+:$U$ is a [[Definition:Subset|subset]] of $S$.
+In a case when $y = 0$:
+:$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$
+By definition of [[Definition:Subset|subset]]:
+:$\map {B_r} {x, r} \subseteq S$ and $\set {\tuple {x, 0} } \subseteq S$
+By [[Union is Smallest Superset]]:
+:$U$ is a [[Definition:Subset|subset]] of $S$.
+Ad $(\text N 1)$:
+Let $\tuple {x, y} \in S$.
+Let $U \in \map \BB {x, y}$.
+In a case when $y > 0$ by $(2)$:
+:$\exists r > 0: U = \map {B_r} {x, y} \cap S$
+By [[Definition:Metric Space Axioms|metric spce axioms]]:
+:$\map d {\tuple {x, y}, \tuple {x, y} } = 0$
+where $d$ denotes the [[Definition:Distance Function|distance function]] of [[Definition:Real Euclidean Space|$\R^2$ Euclidean space]].
+By definition of [[Definition:Open Ball|open ball]]:
+:$\tuple {x, y} \in \map {B_r} {x, y}$
+Thus by definition of [[Definition:Set Intersection|intersection]]:
+:$\tuple {x, y} \in U$
+In a case when $y = 0$ by $(3)$:
+:$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$
+By definition of [[Definition:Singleton|singleton]]:
+:$\tuple {x, 0} \in \set {\tuple {x, 0} }$
+Thus by definition of [[Definition:Set Union|union]]:
+:$\tuple {x, 0} \in U$
+Thus in both cases:
+:$\tuple {x, 0} \in U$
+Ad $(\text N 2)$:
+Let $\tuple {x, 0} \in S, U \in \map \BB {x, y}, \tuple {z, s} \in U$.
+In a case when $y > 0$ by $(2)$:
+:$\exists r > 0: U = \map {B_r} {x, y} \cap S$
+By definition of [[Definition:Set Intersection|intersection]]:
+:$\tuple {z, s} \in \map {B_r} {x, y}$
+By definition of [[Definition:Open Ball|open ball]]:
+:$\map d {\tuple {x, y}, \tuple {z, s} } < r$
+In a subcase when $s > 0$ define
+:$r' := r - \map d {\tuple {x, y}, \tuple {z, s} }$
+and
+:$V := \map {B_{r'} } {z, s} \cap S$
+By $(2)$:
+:$V \in \map \BB {z, s}$
+By proof of [[Open Ball of Point Inside Open Ball]]:
+:$\map {B_{r'} } {z, s} \subseteq \map {B_r} {x, y}$
+Thus by [[Set Intersection Preserves Subsets/Corollary]]:
+:$V \subseteq U$
+In a subcase when $s = 0$ define
+:$r'' := \dfrac {r - \map d {\tuple {x, y}, \tuple {z, s} } } 2$
+and
+:$V := \map {B_{r''} } {z, r''} \cup \set {\tuple {z, 0} }$
+By $(3)$:
+:$V \in \map \BB {z, s}$
+{{begin-eqn}}
+{{eqn | l = \map d {\tuple {x, y}, \tuple {z, r''} }
+      | o = \le
+      | r = \map d {\tuple {x, y}, \tuple {z, 0} } + \map d {\tuple {z, 0}, \tuple {z, r''} }
+      | c = {{Defof|Metric Space Axioms}}
+}}
+{{eqn | r = \map d {\tuple {x, y}, \tuple {z, 0} } + r''
+      | c = {{Defof|Real Number Plane with Euclidean Metric}}
+}}
+{{eqn | r = r - r''}}
+{{end-eqn}}
+By [[Open Ball is Subset of Open Ball]]:
+:$\map {B_{r''} } {z, r''} \subseteq \map {B_r} {x, y}$
+and
+:$\map {B_{r''} } {z, r''} \subseteq S$
+By [[Intersection is Largest Subset]]:
+:$\map {B_{r''} } {z, r''} \subseteq U$
+By definition of [[Definition:Subset|subset]]:
+:$\set {\tuple {z, 0} } \subseteq U$
+Thus by [[Union is Smallest Superset]]:
+:$V \subseteq U$
+In a case when $y = 0$ by $(3)$:
+:$\exists r > 0: U = \map {B_r} {x, r} \cup \set {\tuple {x, 0} }$
+By definition of [[Definition:Set Union|onion]] and [[Definition:Singleton|singleton]]:
+:$\tuple {z, s} \in \map {B_r} {x, r}$ or $\tuple {z, s} = \tuple {x, 0}$
+In a subcase when $\tuple {z, s} = \tuple {x, 0}$ define:
+:$V := U$
+By $(3)$:
+:$V \in \map \BB {z, s}$
+Thus by [[Set is Subset of Itself]]:
+:$V \subseteq U$
+In a subcase when $\tuple {z, s} \in \map {B_r} {x, r}$, define:
+:$r' := r - \map d {\tuple {x, r}, \tuple {z, s} }$
+and
+:$V := \map {B_{r'} } {z, s} \cap S$
+By definition of [[Definition:Open Ball|open ball]]:
+:$\map d {\tuple {x, r}, \tuple {z, s} } < r$
+Then $r' > 0$
+By $(2)$:
+:$V \in \map \BB {z, s}$
+By [[Open Ball of Point Inside Open Ball]]:
+:$\map {B_{r'} } {z, s} \subseteq \map {B_{r'} } {x, r}$
+By [[Intersection is Subset]]:
+:$V = \map {B_{r'} } {z, s} \cap S \subseteq \map {B_{r'} } {z, s}$
+By [[Set is Subset of Union]]:
+:$\map {B_{r'} } {x, r} \subseteq U$
+Thus by [[Subset Relation is Transitive]]:
+$V \subseteq U$
+Ad $(\text N 3)$:
+Let $\tuple {x, y} \in S, U_1, U_2 \in \map \BB {x, y}$.
+In a case when $y > 0$ by $(2)$:
+:$\exists r_1 > 0: U_1 = \map {B_{r_1} } {x, y} \cap S$
+and
+:$\exists r_2 > 0: U_2 = \map {B_{r_2} } {x, y} \cap S$
+By [[Trichotomy Law for Real Numbers]]:
+:$r_1 \le r_2$ or $r_1 \ge r_2$
+{{WLOG}}, suppose 
+:$r_1 \le r_2$
+{{begin-eqn}}
+{{eqn | l = r_2 - r_1
+      | o = \ge
+      | r = 0
+}}
+{{eqn | r = \map d {\tuple {x, y}, \tuple {x, y} }
+      | c = [[Definition:Metric Space Axioms|metric space axioms]]
+}}
+{{end-eqn}}
+Then by [[Open Ball is Subset of Open Ball]]:
+:$\map {B_{r_1} } {x, y} \subseteq \map {B_{r_2} } {x, y}$
+By [[Set Intersection Preserves Subsets/Corollary]]:
+:$U_1 \subseteq U_2$
+By [[Intersection with Subset is Subset]]:
+:$U_1 = U_1 \cap U_2$
+Thus by [[Set is Subset of Itself]]:
+:$U_1 \subseteq U_1 \cap U_2$
+In a case when $y = 0$ by $(3)$:
+:$\exists r_1 > 0: U_1 = \map {B_{r_1} } {x, r_1} \cup \set {\tuple {x, 0} }$
+and
+:$\exists r_2 > 0: U_2 = \map {B_{r_2} } {x, r_2} \cup \set {\tuple {x, 0} }$
+By [[Trichotomy Law for Real Numbers]]:
+:$r_1 \le r_2$ or $r_1 \ge r_2$
+{{WLOG}}, suppose 
+:$r_1 \le r_2$
+{{begin-eqn}}
+{{eqn | l = r_2 - r_1
+      | r = \map d {\tuple {x, r_2}, \tuple {x, r_1} }
+      | c = {{Defof|Real Number Plane with Euclidean Metric}}
+}}
+{{end-eqn}}
+Then by [[Open Ball is Subset of Open Ball]]:
+:$\map {B_{r_1} } {x, r_1} \subseteq \map {B_{r_2} } {x, r_2}$
+By [[Set Union Preserves Subsets/Corollary]]:
+:$U_1 \subseteq U_2$
+By [[Intersection with Subset is Subset]]:
+:$U_1 = U_1 \cap U_2$
+Thus by [[Set is Subset of Itself]]:
+:$U_1 \subseteq U_1 \cap U_2$
+Thus by $(\text N 0)$-$(\text N 3)$, $(4)$ and [[Topology Defined by Neighborhood System]]:
+:$T = \struct {S, \tau}$ is a [[Definition:Topological Space|topological space]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Solution of Second Order Differential Equation with Missing Independent Variable}
+Tags: Second Order ODEs
+
+\begin{theorem}
+Let $\map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$ be a [[Definition:Second Order Ordinary Differential Equation|second order ordinary differential equation]] in which the [[Definition:Independent Variable|independent variable]] $x$ is not explicitly present.
+Then $g$ can be reduced to a [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]], whose [[Definition:General Solution|solution]] can be determined.
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Second Order Ordinary Differential Equation|second order ordinary differential equation]]:
+:$(1): \quad \map g {y, \dfrac {\d y} {\d x}, \dfrac {\d^2 y} {\d x^2} } = 0$
+Let a new [[Definition:Dependent Variable|dependent variable]] $p$ be introduced:
+:$y' = p$
+Hence:
+:$y'' = \dfrac {\d p} {\d x} = \dfrac {\d p} {\d y} \dfrac {\d y} {\d x} = p \dfrac {\d p} {\d y}$
+Then $(1)$ can be transformed into:
+:$(2): \quad \map g {y, p, p \dfrac {\d p} {\d y} = 0}$
+which is a [[Definition:First Order ODE|first order ODE]].
+If $(2)$ has a [[Definition:General Solution|solution]] which can readily be found, it will be expressible in the form:
+:$(3): \quad \map g {x, p}$
+which can then be expressed in the form:
+:$\map g {x, \dfrac {\d y} {\d x} }$
+which is likewise subject to the techniques of solution of a [[Definition:First Order ODE|first order ODE]].
+Hence such a [[Definition:Second Order ODE|second order ODE]] is reduced to the problem of solving two [[Definition:First Order ODE|first order ODEs]] in succession.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + k^2 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs, Linear Second Order ODE: y'' + k^2 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + k^2 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = A \, \map \sin {k x + B}$
+or can be expressed as:
+:$y = C_1 \sin k x + C_2 \cos k x$
+\end{theorem}<|endoftext|>
+\section{Second Order ODE/y y'' + (y')^2 = 0}
+Tags: Examples of Second Order ODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y y'' + \paren {y'}^2 = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y^2 = C_1 x + C_2$
+\end{theorem}
+
+\begin{proof}
+Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as:
+{{begin-eqn}}
+{{eqn | l = y p \frac {\d p} {\d y} + p^2
+      | r = 0
+      | c = where $p = \dfrac {\d y} {\d x}$
+}}
+{{eqn | ll= \leadsto
+      | l = y \rd p + p \rd y
+      | r = 0
+      | c = multiplying by $\dfrac {\d y} p$
+}}
+{{eqn | ll= \leadsto
+      | l = p y
+      | r = C
+      | c = [[First Order ODE/y dx + x dy = 0|First Order ODE: $y \rd x + x \rd y = 0$]]
+}}
+{{eqn | ll= \leadsto
+      | l = y \dfrac {\d y} {\d x}
+      | r = C
+      | c = substituting $p = \dfrac {\d y} {\d x}$
+}}
+{{eqn | ll= \leadsto
+      | l = y^2
+      | r = 2 C x + C_2
+      | c = [[First Order ODE/y dy = k dx|First Order ODE: $y \dfrac {\d y} {\d x} = k$]]
+}}
+{{eqn | r = C_1 x + C_2
+      | c = reassigning constant
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/x y'' = y' + (y')^3}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x y'' = y' + \paren {y'}^3$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x^2 + \paren {y - C_2}^2 = C_1^2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]].
+Substitute $p$ for $y'$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = x \dfrac {\d p} {\d x}
+      | r = p + p^3
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p = \frac {\d y} {\d x}
+      | r = \frac x {\sqrt {C_1^2 - x^2} }
+      | c = [[First Order ODE/x dy = (y + y^3) dx|First Order ODE: $x \rd y = \paren {y + y^3} \rd x$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \rd y
+      | r = \int \frac x {\sqrt {C_1^2 - x^2} }
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = -\sqrt {C_1^2 - x^2} + C_2
+      | c = [[Primitive of x over Root of a squared minus x squared|Primitive of $\dfrac x {\sqrt{a^2 - x^2} }$]]
+}}
+{{eqn | ll= \leadsto
+      | l = x^2 + \paren {y - C_2}^2
+      | r = C_1^2
+      | c = rearranging
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - k^2 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs, Linear Second Order ODE: y'' - k^2 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - k^2 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 e^{k x} + C_2 e^{-k x}$
+\end{theorem}<|endoftext|>
+\section{Second Order ODE/x^2 y'' = 2 x y' + (y')^2}
+Tags: Examples of Second Order ODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$x^2 y'' = 2 x y' + \paren {y'}^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = -\dfrac {x^2} 2 - C_1 x - {C_1}^2 \, \map \ln {x - C_1} + C_2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]].
+Substitute $p$ for $y'$:
+{{begin-eqn}}
+{{eqn | l = x^2 \dfrac {\d p} {\d x} 
+      | r = 2 x p + p^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p = \frac {\d y}{\d x}
+      | r = -\frac {x^2} {x - C_1}
+      | c = [[Bernoulli's Equation/x^2 dy = (2 x y + y^2) dx|Bernoulli's Equation: $x^2 \rd y = \paren {2 x y + y^2} \rd x$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \rd y
+      | r = -\int \frac {x^2} {x - C_1} \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = -\frac {x^2} 2 - C_1 x - {C_1}^2 \, \map \ln {x - C_1} + C_2
+      | c = [[Primitive of x squared over a x + b|Primitive of $\dfrac {x^2} {a x + b}$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/(x^2 + 2 y') y'' + 2 x y' = 0}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad \paren {x^2 + 2 y'} y'' + 2 x y' = 0$
+subject to the [[Definition:Initial Condition|initial conditions]]:
+:$y = 1$ and $y' = 0$ when $x = 0$
+has the [[Definition:Particular Solution of Differential Equation|particular solution]]:
+:$y = 1$
+or:
+:$3 y + x^3 = 3$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]].
+Substitute $p$ for $y'$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = \paren {x^2 + 2 p} \dfrac {\d p} {\d x} + 2 x p
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 x p \rd x + \paren {x^2 + 2 p} \rd p 
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = p \paren {x^2 + p}
+      | r = C_1 
+      | c = [[Bernoulli's Equation/2 x y dx + (x^2 + 2 y) dy = 0|Bernoulli's Equation: $2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$]]
+}}
+{{end-eqn}}
+Consider the [[Definition:Initial Condition|initial condition]]:
+:$y' = p = 0$ when $x = 0$
+Hence putting $p = x = 0$ in $(2)$ we get:
+:$0 \cdot 0^2 + 0^2 = C_1$
+:$C_1 = 0$
+and so $(2)$ becomes:
+{{begin-eqn}}
+{{eqn | l = p x^2
+      | r = -p^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p \paren {x^2 - p}
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+There are two possibilities here:
+{{begin-eqn}}
+{{eqn | l = p
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d y} {\d x}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = C_2
+      | c = 
+}}
+{{end-eqn}}
+From our [[Definition:Initial Condition|initial condition]]:
+:$y = 1$ when $x = 0$
+gives us:
+:$C_2 = 1$
+and so the [[Definition:Particular Solution of Differential Equation|solution]] is obtained:
+:$y = 1$
+{{qed|lemma}}
+The other option is:
+{{begin-eqn}}
+{{eqn | l = p = \dfrac {\d y} {\d x}
+      | r = -x^2 
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = -\int x^2 \rd x
+      | c = 
+}}
+{{eqn | n = 3
+      | r = -\frac {x^3} 3 + C_2
+      | c = 
+}}
+{{end-eqn}}
+From our [[Definition:Initial Condition|initial condition]]:
+:$y = 1$ when $x = 0$
+Hence putting $x = 0$ and $y = 1$ in $(3)$ we get:
+:$1 = - \dfrac {0^3} 3 = C_2$
+and so $C_2 = 1$.
+Thus we have:
+:$y + \dfrac {x^3} 3 = 1$
+or:
+:$3 y + x^3 = 3$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/y y'' = y^2 y' + (y')^2}
+Tags: Examples of Second Order ODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y y'' = y^2 y' + \paren {y'}^2$
+subject to the [[Definition:Initial Condition|initial conditions]]:
+:$y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$
+has the [[Definition:Particular Solution of Differential Equation|particular solution]]:
+:$2 y - 3 = 8 y \, \map \exp {\dfrac {3 x} 2}$
+\end{theorem}
+
+\begin{proof}
+Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as:
+{{begin-eqn}}
+{{eqn | l = y p \frac {\d p} {\d y}
+      | r = y^2 p + p^2
+      | c = where $p = \dfrac {\d y} {\d x}$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d p} {\d y} - \frac p y
+      | r = y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p = \dfrac {\d y} {\d x}
+      | r = y \paren {y + C_1}
+      | c = [[Linear First Order ODE/y' - (y over x) = k x|Linear First Order ODE: $y' - \dfrac y x = k x$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d y} {y \paren {y + C_1} }
+      | r = \int \rd x
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {C_1} \map \ln {\frac y {y + C_1} }
+      | r = x + C_2
+      | c = [[Primitive of Reciprocal of x by a x + b|Primitive of Reciprocal of $\dfrac x {\paren {a x + b} }$]]
+}}
+{{end-eqn}}
+Now to consider the [[Definition:Initial Condition|initial conditions]]:
+:$y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$
+After algebra:
+{{begin-eqn}}
+{{eqn | l = \map \ln {\frac y {y + C_1} }
+      | r = C_1 x + C_2
+      | c = reassigning $C_2$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac y {y + C_1}
+      | r = e^{C_1 x + C_2}
+      | c = 
+}}
+{{eqn | r = e^{C_2} e^{C_1 x}
+      | c = 
+}}
+{{end-eqn}}
+When $x = 0$ we have $y = -1/2$:
+{{begin-eqn}}
+{{eqn | l = \frac {-1/2} {-1/2 + C_1}
+      | r = \frac 1 {1 - 2 C_1}
+      | c = 
+}}
+{{eqn | r = e^{C_2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac y {y + C_1}
+      | r = \frac {e^{C_1 x} } {1 - 2 C_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y \paren {1 - 2 C_1}
+      | r = \paren {y + C_1} e^{C_1 x}
+      | c = 
+}}
+{{end-eqn}}
+Differentiating to get $y'$:
+:$y' \paren {1 - 2 C_1} = \paren {y + C_1} C_1 e^{C_1 x} + e^{C_1 x} y'$
+Putting $y' = 1$ when $x = 0$ we get:
+{{begin-eqn}}
+{{eqn | l = 1 - 2 C_1
+      | r = \paren {-\frac 1 2 + C_1} C_1 + 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_1
+      | r = -\frac 3 2
+      | c = 
+}}
+{{end-eqn}}
+So:
+{{begin-eqn}}
+{{eqn | l = y \paren {1 - 2 C_1}
+      | r = \paren {y + C_1} e^{C_1 x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y \paren {1 - 2 \frac {-3} 2}
+      | r = \paren {y - \frac 3 2} e^{\frac {-3 x} 2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 8 y
+      | r = \paren {2 y - 3} e^{\frac {-3 x} 2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 y - 3
+      | r = 8 y \, \map \exp {\dfrac {3 x} 2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/y'' = 1 + (y')^2}
+Tags: Examples of Second Order ODEs, Second Order ODE: y'' = 1 + (y')^2
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' = 1 + \paren {y'}^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \map {\ln \sec} {x + C_1} + C_2$
+\end{theorem}<|endoftext|>
+\section{Cycloid has Tautochrone Property}
+Tags: Cycloids
+
+\begin{theorem}
+Consider a [[Definition:Wire|wire]] bent into the shape of an [[Definition:Arc of Cycloid|arc]] of a [[Definition:Cycloid|cycloid]] $C$ and inverted so that its [[Definition:Cusp of Cycloid|cusps]] are uppermost and on the same [[Definition:Horizontal Line|horizontal line]].
+Let a [[Definition:Bead|bead]] $B$ be released from some [[Definition:Point|point]] on the [[Definition:Wire|wire]].
+The [[Definition:Time|time]] taken for $B$ to reach the lowest [[Definition:Point|point]] of $C$ is:
+:$T = \pi \sqrt {\dfrac a g}$
+independently of the [[Definition:Point|point]] at which $B$ is released from.
+That is, a [[Definition:Cycloid|cycloid]] is a [[Definition:Tautochrone|tautochrone]].
+\end{theorem}
+
+\begin{proof}
+:[[File:Brachistochrone.png|500px]]
+By the [[Principle of Conservation of Energy]], the [[Definition:Speed|speed]] of the bead at a particular height is determined by its loss in [[Definition:Potential Energy|potential energy]] in getting there.
+Thus, at the [[Definition:Point|point]] $\tuple {x, y}$, we have:
+:$(1): \quad v = \dfrac {\d s} {\d t} = \sqrt {2 g y}$
+This can be written:
+{{begin-eqn}}
+{{eqn | l = \d t
+      | r = \frac {\d s} {\sqrt {2 g y} }
+      | c = 
+}}
+{{eqn | r = \frac {\sqrt {\d x^2 + \d y^2} } {\sqrt {2 g y} }
+      | c = 
+}}
+{{end-eqn}}
+Thus the [[Definition:Time|time]] taken for the bead to slide down the wire is given by:
+:$\displaystyle T_1 = \int \sqrt {\dfrac {\d x^2 + \d y^2} {2 g y} }$
+From [[Equation of Cycloid]], we have:
+:$x = a \paren {\theta - \sin \theta}$
+:$y = a \paren {1 - \cos \theta}$
+Substituting these in the above integral:
+:$\displaystyle T_1 = \int_0^{\theta_1} \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {1 - \cos \theta} } } \rd \theta = \theta_1 \sqrt {\dfrac a g}$
+This is the time needed for the bead to reach the bottom when released when $\theta_1 = \pi$, and so:
+:$T_1 = \pi \sqrt {\dfrac a g}$
+Now suppose the bead is released at any intermediate point $\tuple {x_0, y_0}$.
+Take equation $(1)$ and replace it with:
+:$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$
+Thus the total time to reach the bottom is:
+{{begin-eqn}}
+{{eqn | l = T
+      | r = \int_{\theta_0}^\pi \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {\cos \theta_0 - \cos \theta} } } \rd \theta
+      | c = 
+}}
+{{eqn | r = \sqrt {\dfrac a g} \int_{\theta_0}^\pi \sqrt {\dfrac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta
+      | c = 
+}}
+{{eqn | n = 2
+      | r = \sqrt {\dfrac a g} \int_{\theta_0}^\pi \dfrac {\sin \frac 1 2 \theta \rd \theta} {\sqrt {\cos^2 \frac 1 2 \theta_0 - \cos^2 \frac 1 2 \theta} } \rd \theta
+      | c = [[Double Angle Formula for Cosine]]
+}}
+{{end-eqn}}
+Setting:
+:$u = \dfrac {\cos \frac 1 2 \theta} {\cos \frac 1 2 \theta_0}$
+and so:
+:$\d u = -\dfrac 1 2 \dfrac {\sin \frac 1 2 \theta \rd \theta} {\cos \frac 1 2 \theta_0}$
+Then $(2)$ becomes:
+{{begin-eqn}}
+{{eqn | l = T
+      | r = -2 \sqrt {\dfrac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }
+      | c = 
+}}
+{{eqn | r = 2 \sqrt {\dfrac a g} \Big [{\arcsin u}\Big]_0^1
+      | c = 
+}}
+{{eqn | r = \pi \sqrt {\dfrac a g}
+      | c = 
+}}
+{{end-eqn}}
+That is, wherever the bead is released from, it takes that same time to reach the bottom.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Velocity of Bead on Brachistochrone}
+Tags: Cycloids
+
+\begin{theorem}
+Consider a [[Definition:Wire|wire]] bent into the shape of an [[Definition:Arc of Cycloid|arc]] of a [[Definition:Cycloid|cycloid]] $C$ and inverted so that its [[Definition:Cusp of Cycloid|cusps]] are uppermost and on the same [[Definition:Horizontal Line|horizontal line]].
+Let $C$ be defined by [[Equation of Cycloid]] embedded in a [[Definition:Cartesian Plane|cartesian plane]]:
+:$x = a \left({\theta - \sin \theta}\right)$
+:$y = -a \left({1 - \cos \theta}\right)$
+Let a [[Definition:Bead|bead]] $B$ be released from some [[Definition:Point|point]] on the [[Definition:Wire|wire]].
+Let $B$ slide without [[Definition:Friction|friction]] under the influence of a constant [[Definition:Gravitational Field|gravitational field]] exerting an [[Acceleration Due to Gravity|acceleration]] $g$.
+Let $s_0$ be the [[Definition:Arc Length|arc length]] along the [[Definition:Cycloid|cycloid]].
+Let $s$ be the [[Definition:Arc Length|arc length]] along the [[Definition:Cycloid|cycloid]] at any subsequent point in [[Definition:Time|time]].
+Then the [[Definition:Speed|speed]] $v$ of $B$ relative to $C$ is defined by the equation:
+:$4 a v^2 = g \left({ {s_0}^2 - s^2}\right)$
+\end{theorem}
+
+\begin{proof}
+By [[Brachistochrone is Cycloid]], $C$ is a [[Definition:Brachistochrone|brachistochrone]].
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Open Ball is Subset of Open Ball}
+Tags: Open Balls
+
+\begin{theorem}
+Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
+Let $x, y$ be [[Definition:Element|points]] of $A$.
+Then:
+: $\epsilon-\delta \ge \map d {x, y} \implies \map {B_\delta} y \subseteq \map {B_\epsilon} x$
+where $\map {B_\epsilon} x$ denotes the [[Definition:Open Ball|open $\epsilon$-ball]] in $M = \struct {A, d}$.
+\end{theorem}
+
+\begin{proof}
+Let $\epsilon - \delta \ge \map d {x, y}$.
+Then $\epsilon \ge \map d {x, y} + \delta$.
+If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$.
+So:
+{{begin-eqn}}
+{{eqn | l = \map d {x, z}
+      | o = \le
+      | r = \map d {x, y} + \map d {y, z}
+      | c = [[Definition:Metric Space Axioms|metric space axioms]]
+}}
+{{eqn | o = <
+      | r = \map d {x, y} + \delta
+      | c = {{Defof|Open Ball}}
+}}
+{{eqn | o = \le
+      | r = \epsilon
+}}
+{{end-eqn}}
+Thus $z \in \map {B_\epsilon} x$.
+So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pursuit Curve of Boat in River}
+Tags: Pursuit Curves
+
+\begin{theorem}
+Consider a straight river $R$ whose [[Definition:Parallel Lines|parallel]] banks are aligned with the [[Definition:Y-Axis|$y$-axis]] and the [[Definition:Straight Line|line]] $x = c$ of a [[Definition:Cartesian Plane|cartesian plane]].
+Let the current of $R$ have a constant and uniform [[Definition:Speed|speed]] $a$ in the [[Definition:Negative Real Number|negative]] $y$ direction.
+Let a boat $B$ be launched from the point $\left({c, 0}\right)$ and headed directly towards the [[Definition:Origin|origin]] with [[Definition:Speed|speed]] $b$ relative to the water.
+The path of $B$ is defined by the equation:
+:$c^k \left({y + \sqrt {x^2 + x^2} }\right) = x^{k + 1}$
+\end{theorem}
+
+\begin{proof}
+The [[Definition:Vector Component|components]] of the [[Definition:Velocity|velocity]] of $b$ are:
+:$\dfrac {\mathrm d x} {\mathrm d t} = - b \cos \theta$
+:$\dfrac {\mathrm d y} {\mathrm d t} = - a + b \sin \theta$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \frac {\mathrm d y} {\mathrm d x}
+      | r = \frac {-a + b \sin \theta} {-b \cos \theta}
+      | c = 
+}}
+{{eqn | r = \frac {-a + b \left({-\dfrac y {\sqrt {x^2 + y^2} } }\right)} {-b \left({\dfrac x {\sqrt {x^2 + y^2} } }\right)}
+      | c = 
+}}
+{{eqn | n = 1
+      | r = \frac {-a \sqrt {x^2 + y^2} + b y} {b x}
+      | c = 
+}}
+{{end-eqn}}
+$(1)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+Let $z = \dfrac y x$.
+Then by [[Solution to Homogeneous Differential Equation]]:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\mathrm d z} {\dfrac {-a \sqrt {1 + z^2} + b z} b - z} + C
+      | c = 
+}}
+{{eqn | r = \int \frac {b \, \mathrm d z} {-a \sqrt {1 + z^2} + b z - b z} + C
+      | c = 
+}}
+{{eqn | r = \int \frac {b \, \mathrm d z} {-a \sqrt {1 + z^2} } + C
+      | c = 
+}}
+{{eqn | r = -\frac b a \int \frac {\mathrm d z} {\sqrt {1 + z^2} } + C
+      | c = 
+}}
+{{eqn | r = -\frac b a \ln \left({z + \sqrt {1 + z^2} }\right)  + C
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared|Primitive of $\dfrac 1 {\sqrt{x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+Substituting back for $z$ and rearranging:
+:$C^k \left({y + \sqrt {x^2 + x^2} }\right) = x^{k + 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R, C in Series}
+Tags: Electronics
+
+\begin{theorem}
+Consider the electrical circuit $K$ consisting of:
+: a [[Definition:Resistance|resistance]] $R$
+: an [[Definition:Inductance|inductance]] $L$
+: a [[Definition:Capacitance|capacitance]] $C$
+in [[Definition:Series (Electronics)|series]] with a source of [[Definition:Electromotive Force|electromotive force]] $E$ which is a [[Definition:Function|function]] of [[Definition:Time|time]] $t$.
+:[[File:CircuitRLCseries.png]]
+The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the [[Definition:Linear Second Order ODE|linear second order ODE]]:
+:$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} + \dfrac 1 C I = \dfrac {\mathrm d E} {\mathrm d t}$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$E_L$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $L$
+:$E_R$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $R$
+:$E_C$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $C$.
+From [[Kirchhoff's Voltage Law]]:
+:$E - E_L - E_R - E_C = 0$
+From [[Ohm's Law]]:
+:$E_R = R I$
+From [[Drop in EMF caused by Inductance is proportional to Rate of Change of Current]]:
+:$E_L = L \dfrac {\mathrm d I} {\mathrm d t}$
+From [[Drop in EMF caused by Capacitance is proportional to Accumulated Charge]]:
+:$E_C = \dfrac 1 C Q$
+where $Q$ is the [[Definition:Electric Charge|electric charge]] $Q$ that has accumulated on $C$.
+Thus:
+:$E - L \dfrac {\mathrm d I} {\mathrm d t} - R I - \dfrac 1 C Q = 0$
+which can be rewritten:
+:$L \dfrac {\mathrm d I} {\mathrm d t} + R I + \dfrac 1 C Q = E$
+[[Definition:Differentiation|Differentiating]] {{WRT|Differentiation}} $t$ gives:
+:$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} + \dfrac 1 C I = \dfrac {\mathrm d E} {\mathrm d t}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Electric Charge in Electric Circuit/L, R, C in Series}
+Tags: Electronics
+
+\begin{theorem}
+Consider the electrical circuit $K$ consisting of:
+: a [[Definition:Resistance|resistance]] $R$
+: an [[Definition:Inductance|inductance]] $L$
+: a [[Definition:Capacitance|capacitance]] $C$
+in [[Definition:Series (Electronics)|series]] with a source of [[Definition:Electromotive Force|electromotive force]] $E$ which is a [[Definition:Function|function]] of [[Definition:Time|time]] $t$.
+:[[File:CircuitRLCseries.png]]
+The [[Definition:Electric Charge|electric charge]] $Q$ in $K$ is given by the [[Definition:Linear Second Order ODE|linear second order ODE]]:
+:$L \dfrac {\mathrm d^2 Q} {\mathrm d t^2} + R \dfrac {\mathrm d Q} {\mathrm d t} + \dfrac 1 C Q = E$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$E_L$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $L$
+:$E_R$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $R$
+:$E_C$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $C$.
+From [[Kirchhoff's Voltage Law]]:
+:$E - E_L - E_R - E_C = 0$
+From [[Ohm's Law]]:
+:$E_R = R I$
+From [[Drop in EMF caused by Inductance is proportional to Rate of Change of Current]]:
+:$E_L = L \dfrac {\mathrm d I} {\mathrm d t}$
+From [[Drop in EMF caused by Capacitance is proportional to Accumulated Charge]]:
+:$E_C = \dfrac 1 C Q$
+where $Q$ is the [[Definition:Electric Charge|electric charge]] $Q$ that has accumulated on $C$.
+Thus:
+:$E - L \dfrac {\mathrm d I} {\mathrm d t} - R I - \dfrac 1 C Q = 0$
+which can be rewritten:
+:$L \dfrac {\mathrm d I} {\mathrm d t} + R I + \dfrac 1 C Q = E$
+By definition of [[Definition:Electric Current|electric current]]:
+:$I = \dfrac {\mathrm d Q} {\mathrm d t}$
+and so:
+:$L \dfrac {\mathrm d^2 Q} {\mathrm d t^2} + R \dfrac {\mathrm d Q} {\mathrm d t} + \dfrac 1 C Q = E$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series}
+Tags: Electronics, Decay Equation
+
+\begin{theorem}
+Consider the electrical circuit $K$ consisting of:
+:a [[Definition:Resistance|resistance]] $R$
+:an [[Definition:Inductance|inductance]] $L$
+in [[Definition:Series (Electronics)|series]] with a source of [[Definition:Electromotive Force|electromotive force]] $E$ which is a [[Definition:Function|function]] of [[Definition:Time|time]] $t$.
+:[[File:CircuitRLseries.png]]
+The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$L \dfrac {\d I} {\d t} + R I = E$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$E_L$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $L$
+:$E_R$ be the drop in [[Definition:Electromotive Force|electromotive force]] across $R$
+From [[Kirchhoff's Voltage Law]]:
+:$E - E_L - E_R = 0$
+From [[Ohm's Law]]:
+:$E_R = R I$
+From [[Drop in EMF caused by Inductance is proportional to Rate of Change of Current]]:
+:$E_L = L \dfrac {\d I} {\d t}$
+Thus:
+:$E - L \dfrac {\d I} {\d t} - R I = 0$
+which can be rewritten:
+:$L \dfrac {\d I} {\d t} + R I = E$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Constant EMF at t = 0}
+Tags: Electronics
+
+\begin{theorem}
+Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$.
+Let a constant [[Definition:Electromotive Force|EMF]] $E_0$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$.
+The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equation:
+:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]:
+:$L \dfrac {\d I} {\d t} + R I = E_0$
+defines the behaviour of $I$.
+This can be written as:
+:$(1): \quad \dfrac {\d I} {\d t} = \dfrac R L \paren {\dfrac {E_0} R - I}$
+$(1)$ is in the form:
+:$\dfrac {\d y} {\d x} = k \paren {y_a - y}$
+where:
+:$k \in \R: k > 0$
+:$y = y_0$ at $x = 0$
+This is an example of the [[Decay Equation]], where:
+:$k = \dfrac R L$
+:$y_a = \dfrac {E_0} R$
+:$y_0 = I_0$.
+whose [[Definition:Particular Solution of Differential Equation|particular solution]] is:
+:$y = y_a + \paren {y_0 - y_a} e^{-k x}$
+Hence the [[Definition:Particular Solution of Differential Equation|particular solution]] to $(1)$ is:
+:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 1}
+Tags: Electronics
+
+\begin{theorem}
+Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$.
+Let a constant [[Definition:Electromotive Force|EMF]] $E_0$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$.
+After a sufficiently long time, the [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equation:
+:$E_0  = R I$
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series/Constant EMF at t = 0|Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$]]:
+:$I = \dfrac {E_0} R + \left({I_0 - \dfrac {E_0} R}\right) e^{-R t / L}$
+We have that:
+:$\displaystyle \lim_{t \mathop \to \infty} e^{-R t / L} \to 0$
+and so:
+:$\displaystyle \lim_{t \mathop \to \infty} I \to \dfrac {E_0} R$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 2}
+Tags: Electronics
+
+\begin{theorem}
+Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $0$.
+Let a constant [[Definition:Electromotive Force|EMF]] $E_0$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$.
+The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equation:
+:$I = \dfrac {E_0} R \paren {1 - e^{-R t / L} }$
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series/Constant EMF at t = 0|Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$]]:
+:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
+Setting $I_0 = 0$:
+{{begin-eqn}}
+{{eqn | l = I
+      | r = \dfrac {E_0} R + \paren {0 - \dfrac {E_0} R} e^{-R t / L}
+      | c = 
+}}
+{{eqn | r = \dfrac {E_0} R \paren {1 - e^{-R t / L} }
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Constant EMF at t = 0/Corollary 3}
+Tags: Electronics
+
+\begin{theorem}
+Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$.
+Let  [[Definition:Electromotive Force|EMF]] imposed upon $K$ be zero.
+The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equation:
+:$I = I_0 e^{-R t / L}$
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series/Constant EMF at t = 0|Electric Current in Electric Circuit: L, R in Series: Constant EMF at $t = 0$]]:
+:$I = \dfrac {E_0} R + \paren {I_0 - \dfrac {E_0} R} e^{-R t / L}$
+Setting $E_0 = 0$:
+{{begin-eqn}}
+{{eqn | l = I
+      | r = \dfrac 0 R + \paren {I_0 - \dfrac 0 R} e^{-R t / L}
+      | c = 
+}}
+{{eqn | r = I_0 e^{-R t / L}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Exponentially Decaying EMF at t = 0}
+Tags: Electronics
+
+\begin{theorem}
+Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$.
+Let an [[Definition:Electromotive Force|EMF]] $E$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$ defined by the equation:
+:$E = E_0 e^{-k t}$
+The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equation:
+:$I = \dfrac {E_0} {R - k L} e^{-k t} + \left({I_0 - \dfrac {E_0} {R - k L} }\right) e^{-R t / L}$
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]:
+:$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E_0 e^{-k t}$
+defines the behaviour of $I$.
+This can be written as:
+:$(1): \quad \dfrac {\mathrm d I} {\mathrm d t} + \dfrac R L I = \dfrac {E_0} L  e^{-k t}$
+$(1)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form:
+:$\dfrac {\mathrm d I} {\mathrm d t} + P \left({t}\right) I = Q \left({t}\right)$
+where:
+:$P \left({t}\right) = \dfrac R L$
+:$Q \left({t}\right) = \dfrac {E_0} L  e^{-k t}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int P \left({t}\right) \, \mathrm d t
+      | r = \int \dfrac R L \, \mathrm d t
+      | c = 
+}}
+{{eqn | r = \dfrac {R t} L
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = e^{\int P \, \mathrm d t}
+      | r = e^{R t / L}
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\dfrac {\mathrm d} {\mathrm d t} \left({e^{R t / L} I}\right) = e^{R t / L} \dfrac {E_0} L e^{-k t}$
+and the [[Definition:General Solution|general solution]] becomes:
+:$\displaystyle e^{R t / L} I = \int e^{R t / L} \dfrac {E_0} L e^{-k t} \, \mathrm d t$
+and so:
+{{begin-eqn}}
+{{eqn | l = e^{R t / L} I
+      | r = \frac {E_0} L  \int e^{R t / L} e^{-k t} \, \mathrm d t
+      | c = 
+}}
+{{eqn | r = \frac {E_0} L  \int e^{\left({R / L - k}\right) t} \, \mathrm d t
+      | c = 
+}}
+{{eqn | r = \frac {E_0} {L \left({\frac R L - k}\right)} e^{\left({R / L - k}\right) t} + C
+      | c = 
+}}
+{{eqn | r = \frac {E_0} {R - k L} e^{\left({R / L - k}\right) t} + C
+      | c = 
+}}
+{{end-eqn}}
+When $t = 0$, we have $I = I_0$.
+So:
+{{begin-eqn}}
+{{eqn | l = I_0
+      | r = \frac {E_0} {R - k L} + C
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = C
+      | r = I_0 - \frac {E_0} {R - k L}
+      | c = 
+}}
+{{end-eqn}}
+So:
+:$I e^{R t/ L} = \dfrac {E_0} {R - k L} e^{\left({R / L - k}\right) t} + I_0 - \dfrac {E_0} {R - k L}$
+Multiplying by $e^{\frac {R t} L}$ and tidying up, we get:
+{{begin-eqn}}
+{{eqn | l = I
+      | r = \frac {E_0} {R - k L} e^{- R t / L + R t / L - k t} + I_0 e^{-R t / L} - \dfrac {E_0} {R - k L} e^{-R t / L}
+      | c = 
+}}
+{{eqn | r = \frac {E_0} {R - k L} e^{- k t} + I_0 e^{-R t / L} - \frac {E_0} {R - k L} e^{-R t / L}
+      | c = 
+}}
+{{eqn | r = \frac {E_0} {R - k L} e^{- k t} + \left({I_0 - \frac {E_0} {R - k L} }\right) e^{-R t / L}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Sinusoidal EMF}
+Tags: Electronics
+
+\begin{theorem}
+Let the [[Definition:Electric Current|electric current]] flowing in $K$ at [[Definition:Time|time]] $t = 0$ be $I_0$.
+Let an [[Definition:Electromotive Force|EMF]] $E$ be imposed upon $K$ at [[Definition:Time|time]] $t = 0$ defined by the equation:
+:$E = E_0 \sin \omega t$
+The [[Definition:Electric Current|electric current]] $I$ in $K$ is given by the equation:
+:$I = \dfrac {E_0} {\sqrt {R^2 - L^2 \omega^2} } \sin \left({\omega t - \alpha}\right) + \left({I_0 - \dfrac {E_0 L \omega} {R^2 + L^2 \omega^2} }\right) e^{-R t / L}$
+where $\tan \alpha = \dfrac {L \omega} R$.
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]:
+:$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E_0 \sin \omega t$
+defines the behaviour of $I$.
+This can be written as:
+:$(1): \quad \dfrac {\mathrm d I} {\mathrm d t} + \dfrac R L I = \dfrac {E_0} L \sin \omega t$
+$(1)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form:
+:$\dfrac {\mathrm d I} {\mathrm d t} + P \left({t}\right) I = Q \left({t}\right)$
+where:
+:$P \left({t}\right) = \dfrac R L$
+:$Q \left({t}\right) = \dfrac {E_0} L  e^{-k t}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int P \left({t}\right) \, \mathrm d t
+      | r = \int \dfrac R L \, \mathrm d t
+      | c = 
+}}
+{{eqn | r = \dfrac {R t} L
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = e^{\int P \, \mathrm d t}
+      | r = e^{R t / L}
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\dfrac {\mathrm d} {\mathrm d t} \left({e^{R t / L} I}\right) = e^{R t / L} \dfrac {E_0} L \sin \omega t$
+and so the [[Definition:General Solution|general solution]] becomes:
+{{begin-eqn}}
+{{eqn | l = e^{R t / L} I
+      | r = \frac {E_0} L \int e^{R t / L} \sin \omega t \, \mathrm d t
+      | c = 
+}}
+{{eqn | r = \frac {E_0} L \frac {e^{R t / L} \left({\frac R L \sin \omega t - \omega \cos \omega t}\right)} {\left({\frac R L}\right)^2 + \omega^2} + C
+      | c = [[Primitive of Exponential of a x by Sine of b x|Primitive of $e^{a x} \sin b x$]]
+}}
+{{eqn | r = \frac {E_0 \left({R \sin \omega t - L \omega \cos \omega t}\right)} {R^2 + \omega^2 L^2} e^{R t / L} + C
+      | c = 
+}}
+{{end-eqn}}
+When $t = 0, I = I_0$.
+{{begin-eqn}}
+{{eqn | l = I_0
+      | r = E_0 \, {\frac {-L \omega} {R^2 + \omega^2 L^2} } + C
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = C
+      | r = I_0 + \frac {E_0 L \omega} {R^2 + \omega^2 L^2}
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = I
+      | r = \frac {E_0 \left({R \sin \omega t - L \omega \cos \omega t}\right)} {R^2 + \omega^2 L^2} + \left({I_0 + \frac {E_0 L \omega} {R^2 + \omega^2 L^2} }\right) e^{-R t / L}
+      | c = 
+}}
+{{end-eqn}}
+The result follows by application of [[Multiple of Sine plus Multiple of Cosine/Sine Form|Multiple of Sine plus Multiple of Cosine: Sine Form]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Condition for Ohm's Law}
+Tags: Electronics
+
+\begin{theorem}
+[[Ohm's Law]] is satisfied by $K$ whenever the [[Definition:Electric Current|current]] $I$ is at a [[Definition:Local Maximum|maximum]] or a [[Definition:Local Minimum|minimum]].
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]:
+:$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E$
+defines the behaviour of $I$.
+Let $I$ be at a [[Definition:Local Maximum|maximum]] or a [[Definition:Local Minimum|minimum]].
+Then from [[Derivative at Maximum or Minimum]]:
+:$\dfrac {\mathrm d I} {\mathrm d t} = 0$
+and so:
+{{begin-eqn}}
+{{eqn | l = E
+      | r = 0 + R I
+      | c = 
+}}
+{{eqn | r = R I
+      | c = 
+}}
+{{end-eqn}}
+which is [[Ohm's Law]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Minimum Current implies Increasing EMF}
+Tags: Electronics
+
+\begin{theorem}
+Let the [[Definition:Electric Current|current]] $I$ be at a [[Definition:Local Minimum|minimum]].
+Then the [[Definition:Electromotive Force|EMF]] $E$ is [[Definition:Increasing|increasing]].
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]:
+:$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E$
+defines the behaviour of $I$.
+Taking the [[Definition:Derivative|derivative]]:
+:$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} = \dfrac {\mathrm d E} {\mathrm d t}$
+From [[Second Derivative of Real Function at Minimum]]:
+:$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} \ge 0$
+while from [[Derivative at Maximum or Minimum]]:
+:$R \dfrac {\mathrm d I} {\mathrm d t} = 0$
+and so:
+:$\dfrac {\mathrm d E} {\mathrm d t} \ge 0$
+The result follows from [[Increasing Function has Positive Derivative]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Current in Electric Circuit/L, R in Series/Maximum Current implies Decreasing EMF}
+Tags: Electronics
+
+\begin{theorem}
+Let the [[Definition:Electric Current|current]] $I$ be at a [[Definition:Local Maximum|maximum]].
+Then the [[Definition:Electromotive Force|EMF]] $E$ is [[Definition:Decreasing|decreasing]].
+\end{theorem}
+
+\begin{proof}
+From [[Current in Electric Circuit/L, R in Series|Electric Current in Electric Circuit: L, R in Series]]:
+:$L \dfrac {\mathrm d I} {\mathrm d t} + R I = E$
+defines the behaviour of $I$.
+Taking the [[Definition:Derivative|derivative]]:
+:$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} + R \dfrac {\mathrm d I} {\mathrm d t} = \dfrac {\mathrm d E} {\mathrm d t}$
+From [[Second Derivative of Real Function at Minimum]]:
+:$L \dfrac {\mathrm d^2 I} {\mathrm d t^2} \le 0$
+while from [[Derivative at Maximum or Minimum]]:
+:$R \dfrac {\mathrm d I} {\mathrm d t} = 0$
+and so:
+:$\dfrac {\mathrm d E} {\mathrm d t} \le 0$
+The result follows from [[Decreasing Function has Negative Derivative]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/y y'' = (y')^2}
+Tags: Examples of Second Order ODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y y'' = \paren {y'}^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_2 e^{C_1 x}$
+\end{theorem}
+
+\begin{proof}
+Using [[Solution of Second Order Differential Equation with Missing Independent Variable]], $(1)$ can be expressed as:
+{{begin-eqn}}
+{{eqn | l = y p \frac {\d p} {\d y}
+      | r = p^2
+      | c = where $p = \dfrac {\d y} {\d x}$
+}}
+{{eqn | ll= \leadsto
+      | l = y \frac {\d p} {\d y}
+      | r = p
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p = \dfrac {\d y} {\d x}
+      | r = C_1 y
+      | c = [[First Order ODE/x dy = k y dx|First Order ODE: $x \rd y = k y \rd x$]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = C_2 e^{C_1 x}
+      | c = [[First Order ODE/dy = k y dx|First Order ODE: $\d y = k y \rd x$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(1 - x y) y' = y^2}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {1 - x y} y' = y^2$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x y = \ln y + C$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be rearranged as:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d y}
+      | r = \frac {1 - x y} {y^2}
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = \frac {\d x} {\d y} + \frac x y
+      | r = \frac 1 {y^2}
+      | c = 
+}}
+{{end-eqn}}
+It can be seen that $(2)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form:
+:$\dfrac {\d x} {\d y} + \map P y x = \map Q y$
+where:
+:$\map P y = \dfrac 1 y$
+:$\map Q y = \dfrac 1 {y^2}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P y \rd y
+      | r = \int \dfrac 1 y \rd y
+      | c = 
+}}
+{{eqn | r = \ln y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd y}
+      | r = y
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(2)$ can be rewritten as:
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac \d {\d y} } {x y}
+      | r = \dfrac 1 y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x y
+      | r = \int \frac {\d y} y
+      | c = 
+}}
+{{eqn | r = \ln y + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(2 x + 3 y + 1) dx + (2 y - 3 x + 5) dy = 0}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {2 x + 3 y + 1} \rd x + \paren {2 y - 3 x + 5} \rd y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$
+\end{theorem}
+
+\begin{proof}
+Rewriting $(1)$ as:
+:$\dfrac {\d y} {\d x} = -\dfrac {2 x + 3 y + 1} {-3 x + 2 y + 5}$
+we note that it is in the form:
+:$\dfrac {\d y} {\d x} = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$
+where:
+:$a e = -4 \ne b d = 9$
+Hence we can use:
+:[[First Order ODE in form y' = F ((a x + b y + c) over (d x + e y + f))|First Order ODE in form $y' = \map F {\dfrac {a x + b y + c} {d x + e y + f} }$]]
+which can be solved by substituting:
+:$x := z - h$
+:$y := w - k$
+where:
+:$h = \dfrac {c e - b f} {a e - b d}$
+:$k = \dfrac {a f - c d} {a e - b d}$
+So let:
+:$x = z - h$ where $h = \dfrac {\paren {1 \times 2} - \paren {3 \times 5} } {\paren {2 \times 2} - \paren {3 \times -3} } = \dfrac {-13} {13} = -1$
+:$y = w - k$ where $k = \dfrac {\paren {2 \times 5} - \paren {1 \times -3} } {\paren {2 \times 2} - \paren {3 \times -3} } = \dfrac {13} {13} = 1$
+Then $(1)$ is transformed into:
+:$(2): \quad \dfrac {\d w} {\d z} = -\dfrac {2 z + 3 w} {3 z - 2 w}$
+Let:
+:$\map M {w, z} =  2 z + 3 w$
+:$\map N {w, z} =  3 z - 2 w$
+We have that:
+:$\map M {t w, t z} =  2 t z + 3 t w = t \paren {2 z + 3 w} = t \, \map M {x, y}$
+:$\map N {t w, t z} =  3 t z - 2 t w = t \paren {3 z - 2 w} = t \, \map N {x, y}$
+Thus both $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $1$.
+Thus by definition $(2)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+So:
+{{begin-eqn}}
+{{eqn | l = \frac {\d w} {\d z}
+      | r = \frac {2 z + 3 w} {3 z - 2 w}
+      | c = 
+}}
+{{eqn | r = \frac {2 + 3 w / z} {3 - 2 w / z}
+      | c = dividing [[Definition:Numerator|top]] and [[Definition:Denominator|bottom]] by $x$
+}}
+{{eqn | r = \frac {2 + 3 u} {3 - 2 u}
+      | c = substituting $u$ for $w / z$
+}}
+{{end-eqn}}
+By [[Solution to Homogeneous Differential Equation]]:
+:$\displaystyle \ln z = \int \frac {\d u} {\map f {1, u} - u} + C$
+where:
+:$\map f {1, u} = \dfrac {2 + 3 u} {3 - 2 u}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \ln z
+      | r = \int \frac {\d u} {\dfrac {2 + 3 u} {3 - 2 u} - u} + C
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {3 - 2 u} \rd z} {2 + 3 u - u \paren {3 - 2 u} } + C
+      | c = 
+}}
+{{eqn | r = \int \frac {\paren {3 - 2 u} \rd v} {2 \paren {1 + u^2} } + C
+      | c = 
+}}
+{{eqn | r = \dfrac 3 2 \int \dfrac {\d u} {1 + u^2} - \int \dfrac {u \rd u} {1 + u^2} + C
+      | c = 
+}}
+{{eqn | r = \dfrac 3 2 \arctan u - \dfrac 1 2 \, \map \ln {1 + u^2} + C
+      | c = 
+}}
+{{end-eqn}}
+Replacing all the substitutions:
+:$\map \ln {x - 1} = \dfrac 3 2 \, \map \arctan {\dfrac {y + 1} {x - 1} } - \dfrac 1 2 \, \map \ln {1 + \paren {\dfrac {y + 1} {x - 1} } } + C$
+Tidying up and reassigning constants appropriately:
+:$3 \, \map \arctan {\dfrac {y + 1} {x - 1} } = \map \ln {\paren {y + 1}^2 + \paren {x - 1}^2} + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/x y' = Root of (x^2 + y^2)}
+Tags: Examples of Homogeneous Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|solution]]:
+:$3 x^2 \ln x = y \sqrt {x^2 + y^2} + x^2 \, \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2$
+\end{theorem}
+
+\begin{proof}
+We divide through by $x$ to show that $(1)$ is [[Definition:Homogeneous Differential Equation|homogeneous]]:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d y}
+      | r = \frac {\sqrt {x^2 + y^2} } x
+      | c = 
+}}
+{{eqn | r = \sqrt {\frac {x^2 + y^2} {x^2} }
+      | c = 
+}}
+{{eqn | r = \sqrt {1 + \paren {\frac y x}^2}
+      | c = 
+}}
+{{end-eqn}}
+By [[Solution to Homogeneous Differential Equation]], its solution is:
+:$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
+where:
+:$\map f {x, y} = \dfrac {\sqrt {x^2 + y^2} } x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {\sqrt {1 + z^2} - z}
+      | c = 
+}}
+{{eqn | r = \int \paren {\sqrt{1 + z^2} + z} \rd z
+      | c = 
+}}
+{{eqn | r = \frac {z \sqrt {1 + z^2} } 2 + \frac {\map \ln {\sqrt{1 + z^2} + z} } 2 + \frac {z^2} 2 + k
+      | c = [[Primitive of Root of x squared plus a squared|Primitive of $\sqrt {x^2 + a^2}$]]
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \ln x
+      | r = z \sqrt {1 + z^2} + \map \ln {\sqrt{1 + z^2} + z} + z^2 + C
+      | c = where $C = 2 k$
+}}
+{{eqn | r = \frac y x \sqrt {1 + \paren {\frac y x}^2} + \map \ln {\sqrt {1 + \paren {\frac y x}^2} + \frac y x} + \paren {\frac y x}^2 + C
+      | c = 
+}}
+{{eqn | r = \frac y x \sqrt {\frac {x^2 + y^2} {x^2} } + \map \ln {\sqrt {\frac {x^2 + y^2} {x^2} } + \frac y x} + \paren {\frac y x}^2 + C
+      | c = 
+}}
+{{eqn | r = \frac y x \frac {\sqrt {x^2 + y^2} } x + \map \ln {\frac {\sqrt {x^2 + y^2} } x + \frac y x} + \paren {\frac y x}^2 + C
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 x^2 \ln x
+      | r = y \sqrt {x^2 + y^2} + x^2 \, \map \ln {y + \sqrt {x^2 + y^2} } - x^2 \ln x + y^2 + C x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 3 x^2 \ln x
+      | r = y \sqrt {x^2 + y^2} + x^2 \, \map \ln {y + \sqrt {x^2 + y^2} } + y^2 + C x^2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernoulli's Equation/y^2 dx = (x^3 - x y) dy}
+Tags: Examples of Bernoulli's Equation
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad y^2 \rd x = \paren {x^3 - x y} \rd y$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$3 y = 2 x^2 + C x^2 y^2$
+\end{theorem}
+
+\begin{proof}
+Dividing $(1)$ by $y^2$ and rearranging:
+:$(2): \quad \dfrac {\d x} {\d y} + \dfrac x y = \dfrac {x^3} {y^2}$
+It can be seen that $(2)$ is in the form:
+:$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$
+where:
+:$\map P y = \dfrac 1 y$
+:$\map Q y = \dfrac 1 {y^2}$
+:$n = 3$
+and so is an example of [[Definition:Bernoulli's Equation|Bernoulli's equation]].
+By [[Solution to Bernoulli's Equation]] it has the [[Definition:General Solution|general solution]]:
+:$(3): \quad \displaystyle \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$
+where:
+:$\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$
+Thus $\map \mu y$ is evaluated:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - n} \int \map P y \rd y
+      | r = \paren {1 - 3} \int \dfrac 1 y \rd y
+      | c = 
+}}
+{{eqn | r = -2 \ln y
+      | c = 
+}}
+{{eqn | r = \map \ln {y^{-2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \mu y
+      | r = e^{\map \ln {y^{-2} } }
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {y^2}
+      | c = 
+}}
+{{end-eqn}}
+and so substituting into $(3)$:
+{{begin-eqn}}
+{{eqn | l = \frac 1 {y^2} \frac 1 {x^2}
+      | r = \paren {1 - 3} \int \frac 1 {y^2} \frac 1 {y^2} \rd y
+      | c = 
+}}
+{{eqn | r = -2 \int \dfrac {\d y} {y^4}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {y^2 x^2}
+      | r = -\dfrac 2 {-3} \dfrac 1 {y^3} + C
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 3 y
+      | r = 2 x^2 + C x^2 y^2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(x^2 y^3 + y) dx = (x^3 y^2 - x) dy}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {x^2 y^3 + y} \rd x = \paren {x^3 y^2 - x} \rd y$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be expressed as:
+:$(2): \quad \paren {y + x^2 y^3} \rd x + \paren {x - x^3 y^2} \rd y = 0$
+We note that $(2)$ is in the form:
+:$\map M {x, y} \rd x + \map N {x, y} \rd y = 0$
+but is not [[Definition:Exact Differential Equation|exact]].
+So, let:
+:$\map M {x, y} = y + x^2 y^3$
+:$\map N {x, y} = x - x^3 y^2$
+Let:
+:$\map P {x, y} = \dfrac {\partial M} {\partial y} - \dfrac {\partial N} {\partial x}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map P {x, y}
+      | r = \paren {1 + 3 x^2 y^2} - \paren {1 - 3 x^2 y^2}
+      | c = 
+}}
+{{eqn | r = 6 x^2 y^2
+      | c = 
+}}
+{{end-eqn}}
+It can be observed that:
+{{begin-eqn}}
+{{eqn | l = \frac {\map P {x, y} } {\map N {x, y} y - \map M {x, y} x}
+      | r = \frac {6 x^2 y^2} {\paren {x y - x^3 y^3} - \paren {x y + x^3 y^3} }
+      | c = 
+}}
+{{eqn | r = -\frac {6 x^2 y^2} {2 x^3 y^3}
+      | c = 
+}}
+{{eqn | r = -\frac 3 {x y}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\map P {x, y} } {\map M {x, y} }$ is a function of $x y$.
+So [[Integrating Factor for First Order ODE/Function of Product of Variables|Integrating Factor for First Order ODE: Function of Product of Variables]] can be used.
+Let $z = x y$.
+Then:
+:$\map \mu {x y} = \map \mu z = e^{\int \map g z \rd z}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \int \map g z \rd z
+      | r = \int -\frac 3 z \rd z
+      | c = 
+}}
+{{eqn | r = - 3 \ln z
+      | c = 
+}}
+{{eqn | r = \map \ln {z^{-3} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int \map g z \rd z}
+      | r = \frac 1 {z^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {x^3 y^3}
+      | c = 
+}}
+{{end-eqn}}
+Thus an [[Definition:Integrating Factor|integrating factor]] for $(1)$ has been found:
+:$\mu = \dfrac 1 {x^3 y^3}$
+which yields, when multiplying it throughout $(2)$:
+:$\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$
+which is now [[Definition:Exact Differential Equation|exact]].
+By [[First Order ODE/(1 over x^3 y^2 + 1 over x) dx + (1 over x^2 y^3 - 1 over y) dy = 0|First Order ODE: $\paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$]], its [[Definition:General Solution of Differential Equation|general solution]] is:
+:$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/y y'' + (y')^2 - 2 y y' = 0}
+Tags: Examples of Second Order ODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$y y'' + \paren {y'}^2 - 2 y y' = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y^2 = C_2 e^{2 x} + C_1$
+\end{theorem}
+
+\begin{proof}
+Using [[Solution of Second Order Differential Equation with Missing Independent Variable]]:
+{{begin-eqn}}
+{{eqn | l = y p \frac {\d p} {\d y} + p^2 - 2 y p
+      | r = 0
+      | c = where $p = \dfrac {\d y} {\d x}$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d p} {\d y} + \frac p y
+      | r = 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p y
+      | r = y^2 + C
+      | c = [[Linear First Order ODE/y' + (y over x) = k x^n|Linear First Order ODE: $y' + \dfrac y x = k x^n$]]: $k = 2, n = 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d y} {\d x}
+      | r = \frac {y^2 + C} y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int \dfrac {y \rd y} {y^2 + C}
+      | r = \int \d x
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 2 \, \map \ln {y^2 + C}
+      | r = x + k
+      | c = [[Primitive of Function under its Derivative]]
+}}
+{{end-eqn}}
+After algebra, and reassigning constants:
+:$y^2 = C_2 e^{2 x} + C_1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Rationals are Everywhere Dense in Sorgenfrey Line}
+Tags: Sorgenfrey Line
+
+\begin{theorem}
+$\Q$ is [[Definition:Everywhere Dense|everywhere dense]] in the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
+Define:
+:$\BB := \set {\hointr x y: x, y \in \R}$
+where $\hointr x y$ denotes the [[Definition:Right Half-Open Real Interval|right half-open real interval]] between $x$ and $y$.
+By definition of [[Definition:Sorgenfrey Line|Sorgenfrey line]]:
+:$\BB$ is [[Definition:Analytic Basis|basis]] of $T$.
+By definition of [[Definition:Subset|subset]]:
+:$\Q^- \subseteq \R$
+where $\Q^-$ denotes the [[Definition:Closure (Topology)|topological closure]] of $\Q$ in $T$.
+By definition of [[Definition:Set Equality|set equality]] to prove the equality: $\Q^- = \R$, it is necessary to show:
+:$\R \subseteq \Q^-$
+Let $x \in \R$.
+By [[Characterization of Closure by Basis]] it suffices to prove that:
+:$\forall U \in \BB: x \in U \implies U \cap \Q \ne \O$
+Let $U \in \BB$.
+By definition of $\BB$:
+:$\exists y, z \in \R: U = \hointr y z$
+Assume:
+:$x \in U$
+By definition of [[Definition:Half-Open Real Interval|half-open real interval]]:
+:$y \le x < z$
+By [[Between two Real Numbers exists Rational Number]]:
+:$\exists q \in \Q: y < q < z$
+By definition of [[Definition:Half-Open Real Interval|half-open real interval]]:
+:$q \in U$
+Thus by definitions of [[Definition:Set Intersection|intersection]] and [[Definition:Empty Set|empty set]]:
+:$U \cap \Q \ne \O$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/x dy + y dx = x cosine x dx}
+Tags: Examples of Linear First Order ODEs, Linear First Order ODE: x dy + y dx = x cosine x dx
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$x \rd y + y \rd x = x \cos x \rd x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x y = x \sin x + \cos x + C$
+\end{theorem}<|endoftext|>
+\section{First Order ODE/x y dy = x^2 dy + y^2 dx}
+Tags: Examples of First Order ODEs, Examples of Bernoulli's Equation, Examples of Homogeneous Differential Equation, First Order ODE: x y dy = x^2 dy + y^2 dx
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x y \rd y = x^2 \rd y + y^2 \rd x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = x \ln y + C x$
+\end{theorem}<|endoftext|>
+\section{First Order ODE/(exp x - 3 x^2 y^2) y' + y exp x = 2 x y^3}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \paren {e^x - 3 x^2 y^2} y' + y e^x = 2 x y^3$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$y e^x - x^2 y^3 = C$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be expressed as:
+:$\paren {y e^x - 2 x y^3} \rd x + \paren {e^x - 3 x^2 y^2} \rd y = 0$
+Let:
+:$\map M {x, y} = y e^x - 2 x y^3$
+:$\map N {x, y} = e^x - 3 x^2 y^2$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = e^x - 6 x y^2
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = e^x - 6 x y^2
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {y e^x - 2 x y^3} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = y e^x - x^2 y^3 + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {e^x - 3 x^2 y^2} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = y e^x + x^2 y^3 + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = y e^x + x^2 y^3$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$y e^x + x^2 y^3 = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/y'' + 2 x (y')^2 = 0}
+Tags: Examples of Second Order ODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 2 x \paren {y'}^2 = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$C_1 \map \arctan {C_1 x} = y + C_2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]].
+Substitute $p$ for $y'$ in $(1)$ and rearranging:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d p} {\d x}
+      | r = -2 x p^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d p} {p^2}
+      | r = -2 \int x \rd x
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = -\frac 1 p
+      | r = -x^2 + k^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d x} {\d y}
+      | r = x^2 + k^2
+      | c = substituting back for $p$
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d x} {x^2 + k^2} 
+      | r = \int \rd y
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 k \map \arctan {\frac x k}
+      | r = y + C_2
+      | c = [[Primitive of Reciprocal of x squared plus a squared/Arctangent Form|Primitive of $\dfrac 1 {x^2 + a^2}$]]
+}}
+{{eqn | ll= \leadsto
+      | l = C_1 \map \arctan {C_1 x}
+      | r = y + C_2
+      | c = setting $C_1 = \dfrac 1 k$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(y over x^2) dx + (y - 1 over x) dy = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \dfrac y {x^2} \rd x + \paren {y - \dfrac 1 x} \rd y = 0$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$\dfrac {y^2} 2 - \dfrac y x = C$
+\end{theorem}
+
+\begin{proof}
+Let $M$ and $N$ be defined as:
+:$\map M {x, y} = \dfrac y {x^2}$
+:$\map N {x, y} = y - \dfrac 1 x$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\partial M} {\partial y}
+      | r = \frac 1 {x^2}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = \dfrac 1 {x^2}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \dfrac y {x^2} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = - \frac y x + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {y - \dfrac 1 x} \r d y + \map h x
+      | c = 
+}}
+{{eqn | r = \frac {y^2} 2 - \frac y x + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = \dfrac {y^2} 2 - \dfrac y x$
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\dfrac {y^2} 2 - \dfrac y x = C$
+{{qed}}
+</onlyinclude>
+[[Category:Examples of Exact Differential Equation]]
+[[Category:Examples of First Order ODEs]]
+syuvq2v6cpkjnr7awwopekq6nio5baz
+\end{proof}<|endoftext|>
+\section{Focal Property of Parabola}
+Tags: Parabolas
+
+\begin{theorem}
+:[[File:ParabolaFocalProperty.png|500px]]
+\end{theorem}
+
+\begin{proof}
+{{proofread}}
+Consider the following parabola whose axis of symmetry is the y-axis along with an incoming light ray from $B$ which is parallel to the axis of symmetry, reflects off the parabola at the point $P$ and intersects the y-axis at the point $F$. Points $B$, $P$, and $Q$ are co-linear and the point $Q$ also lies on the directrix of the parabola; the line $AT$ is the tangent of the parabola at the point $P$:
+[[File:ParabolaReflection.png|500px]]
+By the [[Law of Reflection]] the angle of incidence, $90 \degrees - \angle BPT$, is equal to the angle of reflection, $90 \degrees - \angle APF$. Hence
+:$\angle BPT = \angle APF$
+Note: Incoming and outgoing light rays both follow the same [[Law of Reflection]]; hence, this argument is valid for both incoming and outgoing light rays.
+From [[Two Straight Lines make Equal Opposite Angles]] we also know that
+:$\angle BPT = \angle APQ$
+We now consider the two triangles $APQ$ and $APF$.
+They have an equal side, $AP$, and an equal angle, $\angle APQ = \angle APF$, and by the [[Triangle Side-Angle-Side Equality]] triangles $APQ$ and $APF$ will be equal if and only if $FP$ and $PQ$.
+From the [[Definition:Parabola/Focus-Directrix|definition of a parabola]] it is the locus of the points that are equidistant from the focus and the directrix; hence the point $F$ is the focus.
+\end{proof}<|endoftext|>
+\section{First Order ODE/(3 x^2 over y^4 - 1 over y^2) dy - 2 x over y^3 dx = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y - \dfrac {2 x} {y^3} \rd x = 0$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equation|general solution]]:
+:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
+\end{theorem}
+
+\begin{proof}
+Let $M$ and $N$ be defined as:
+:$\map M {x, y} = -\dfrac {2 x} {y^3}$
+:$\map N {x, y} = \dfrac {3 x^2} {y^4} - \dfrac 1 {y^2}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\partial M} {\partial y}
+      | r = \frac {6 x} {2 y^4}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = \dfrac {6 x} {y^4}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int - \dfrac {2 x} {y^3} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = -\dfrac {x^2} {y^3} + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {\dfrac {3 x^2} {y^4} - \dfrac 1 {y^2} } \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = -\dfrac {x^2} {y^3} + \dfrac 1 y + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = \dfrac 1 y - \dfrac {x^2} {y^3}$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$ is:
+:$\dfrac 1 y - \dfrac {x^2} {y^3} = C$
+{{qed}}
+[[Category:Examples of Exact Differential Equation]]
+[[Category:Examples of First Order ODEs]]
+hhsdvpmiuh51fgk6yttk6djrk9y3mdj
+\end{proof}<|endoftext|>
+\section{First Order ODE/(y - 1 over x) dx + (x - y) dy = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {y - \dfrac 1 x} \rd x + \paren {x - y} \rd y = 0$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution to Differential Equation|solution]]:
+:$x y - \ln x - \dfrac {y^2} 2 + C$
+\end{theorem}
+
+\begin{proof}
+Let $M$ and $N$ be defined as:
+:$\map M {x, y} = y - \dfrac 1 x$
+:$\map N {x, y} = x - y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\partial M} {\partial y}
+      | r = 1
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution to Differential Equation|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {y - \dfrac 1 x} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = x y - \ln x + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {x - y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = x y - \dfrac {y^2} 2 + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = x y - \ln x - \dfrac {y^2} 2$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution to Differential Equation|solution]] to $(1)$ is:
+:$x y - \ln x - \dfrac {y^2} 2 + C$
+{{qed}}
+[[Category:Examples of Exact Differential Equation]]
+[[Category:Examples of First Order ODEs]]
+pnrufusux3jtoaq69awzlj5ysojqhh3
+\end{proof}<|endoftext|>
+\section{First Order ODE/1 over x^3 y^2 dx + (1 over x^2 y^3 + 3 y) dy = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \dfrac 1 {x^3 y^2} \rd x + \paren {\dfrac 1 {x^2 y^3} + 3 y} \rd y = 0$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
+\end{theorem}
+
+\begin{proof}
+Let $M$ and $N$ be defined as:
+:$\map M {x, y} = \dfrac 1 {x^3 y^2}$
+:$\map N {x, y} = \dfrac 1 {x^2 y^3} + 3 y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\partial M} {\partial y}
+      | r = -\frac 1 {x^3 y^2}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = -\frac 1 {x^3 y^2}
+      | c = 
+}}
+{{end-eqn}}
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \frac 1 {x^3 y^2} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = -\frac 1 {2 x^2 y^2} + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {\frac 1 {x^2 y^3} + 3 y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = -\frac 1 {2 x^2 y^2} + 3 \dfrac {y^2} 2 + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = -\dfrac 2 {x^2 y^2} + \dfrac {3 y^2} 2$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$-\dfrac 1 {2 x^2 y^2} + \dfrac {3 y^2} 2 = C$
+{{qed}}
+[[Category:Examples of Exact Differential Equation]]
+[[Category:Examples of First Order ODEs]]
+h02f2apyn104iuz6ibkxxmdj318amx8
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/y' - (y over x) = 3 x}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = 3 x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\dfrac y x = 3 x + C$
+or:
+:$y = 3 x^2 + C x$
+\end{theorem}
+
+\begin{proof}
+This is an instance of [[Linear First Order ODE/y' - (y over x) = k x|Linear First Order ODE: $y' - \dfrac y x = k x$]].
+Its [[Definition:General Solution of Differential Equation|solution]] is:
+:$\dfrac y x = k x + C$
+or:
+:$y = k x^2 + C x$
+from which the [[Definition:General Solution of Differential Equation|solution]] to $(1)$ is found by substituting $3$ for $k$.
+{{qed}}
+[[Category:Examples of Linear First Order ODEs]]
+5zl5xx9teqim55grzediaigsppbf6nk
+\end{proof}<|endoftext|>
+\section{Characterization of Closure by Basis}
+Tags: Set Closures
+
+\begin{theorem}
+Let $T = \left({S, \tau}\right)$ be a [[definition:Topological Space|topological space]].
+Let $\mathcal B \subseteq \tau$ be a [[Definition:Basis (Topology)/Analytic Basis|basis]].
+Let $A$ be a [[definition:Subset|subset]] of $T$.
+Let $x$ be a point of $T$.
+Then $x \in A^-$ {{iff}}:
+:for every $U \in \mathcal B$:
+::if $x \in U$ then $A \cap U \ne \varnothing$
+where:
+:$A^-$ denotes the [[Definition:Closure (Topology)|closure]] of $A$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $x \in A^-$.
+Let $U \in \mathcal B$.
+By definition of [[Definition:Basis (Topology)/Analytic Basis|basis]], $U$ is an [[Definition:Open Set (Topology)|open set of $T$]].
+Thus from [[Condition for Point being in Closure]]:
+:if $x \in U$ then $A \cap U \ne \varnothing$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $x$ be such that for every $U \in \mathcal B$:
+:if $x \in U$
+:then $A \cap U \ne \varnothing$.
+By [[Condition for Point being in Closure]], to prove that $x \in \operatorname{Fr} A$ it is enough to prove
+that:
+:for every [[Definition:Open Set (Topology)|open set]] $U$ of $T$:
+::if $x \in U$ then $A \cap U \ne \varnothing$.
+Let $U$ be an [[Definition:Open Set (Topology)|open set]] of $T$.
+Let $x \in U$.
+By definition of [[Definition:Basis (Topology)/Analytic Basis|(analytic) basis]], there exists $V \in \mathcal B$ such that:
+:$x \in V \subseteq U$
+By assumption:
+:$A \cap V \ne \varnothing$
+From the [[Set Intersection Preserves Subsets/Corollary|corollary to Set Intersection Preserves Subsets]]:
+:$A \cap V \subseteq A \cap U$
+So:
+:$A \cap U \ne \varnothing$
+and hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sorgenfrey Line is Separable}
+Tags: Sorgenfrey Line
+
+\begin{theorem}
+The [[Definition:Sorgenfrey Line|Sorgenfrey line]] is [[Definition:Separable Space|separable]].
+\end{theorem}
+
+\begin{proof}
+By [[Rationals are Everywhere Dense in Sorgenfrey Line]]:
+:$\Q$ is [[Definition:Everywhere Dense|dense]] in the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
+By [[Rational Numbers are Countably Infinite]]:
+:$\Q$ is [[Definition:Countable Set|countable]].
+Thus by definition:
+:The [[Definition:Sorgenfrey Line|Sorgenfrey line]] is [[Definition:Separable Space|separable]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y dx + x dy = 0}
+Tags: Examples of First Order ODEs, Examples of Separation of Variables
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad y \rd x + x \rd y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$x y = C$
+\end{theorem}
+
+\begin{proof}
+$(1)$ can be expressed as:
+{{begin-eqn}}
+{{eqn | l = x \rd y
+      | r = -y \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = C x^{-1}
+      | c = [[First Order ODE/x dy = k y dx|First Order ODE: $x \rd y = k y \rd x$]]
+}}
+{{eqn | ll= \leadsto
+      | l = x y
+      | r = C
+      | c = multiplying through by $x$
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of First Order ODEs]]
+[[Category:Examples of Separation of Variables]]
+6ezpoyg7g40z9d75edixh9xyydjql0n
+\end{proof}<|endoftext|>
+\section{First Order ODE/y dy = k dx}
+Tags: Examples of First Order ODEs, Examples of Separation of Variables
+
+\begin{theorem}
+Let $k \in \R$ be a [[Definition:Real Number|real number]].
+The [[Definition:First Order ODE|first order ODE]]:
+:$y \, \dfrac {\d y} {\d x} = k$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y^2 = 2 k x + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y \, \dfrac {\d y} {\d x}
+      | r = k
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int y \rd y
+      | r = \int k \rd x
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y^2} 2
+      | r = k x + \frac C 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^2
+      | r = 2 k x + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of First Order ODEs]]
+[[Category:Examples of Separation of Variables]]
+gkokkgsocrjtw5p1968f3m1dbe3iagj
+\end{proof}<|endoftext|>
+\section{First Order ODE/x dy = (y + y^3) dx}
+Tags: Examples of First Order ODEs, Examples of Separation of Variables
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x \dfrac {\d y} {\d x} = y + y^3$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \dfrac x {\sqrt {C^2 - x^2} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x \dfrac {\d y} {\d x}
+      | r = y + y^3
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d y} {y \paren {1 + y^2} }
+      | r = \int \frac {\d x} x
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 2 \, \map \ln {\frac {y^2} {y^2 + 1} }
+      | r = \ln x + \ln k
+      | c = [[Primitive of Reciprocal of x by x squared plus a squared|Primitive of $\dfrac 1 {x \paren {x^2 + a^2} }$]]
+}}
+{{eqn | r = \map \ln {k x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \ln {\frac {y^2} {y^2 + 1} }
+      | r = \map \ln {k^2 x^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y^2} {y^2 + 1} 
+      | r = k^2 x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y^2 + 1} {y^2}
+      | r = \frac 1 {k^2 x^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 1 + \frac 1 {y^2}
+      | r = \frac 1 {k^2 x^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 {y^2}
+      | r = \frac 1 {k^2 x^2} - 1
+      | c = 
+}}
+{{eqn | r = \frac {1 - k^2 x^2} {k^2 x^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^2
+      | r = \frac {k^2 x^2} {1 - k^2 x^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^2
+      | r = \frac {x^2} {\frac 1 {k^2} - x^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^2
+      | r = \frac {x^2} {C^2 - x^2}
+      | c = letting $C = \dfrac 1 k$
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \frac x {\sqrt {C^2 - x^2} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of First Order ODEs]]
+[[Category:Examples of Separation of Variables]]
+gp4lv30s4p8oxy2ivldhtbtf5kgzxi2
+\end{proof}<|endoftext|>
+\section{First Order ODE/y dy = k x dx}
+Tags: Examples of First Order ODEs, Examples of Separation of Variables
+
+\begin{theorem}
+Let $k \in \R$ be a [[Definition:Real Number|real number]].
+The [[Definition:First Order ODE|first order ODE]]:
+:$y \rd y = k x \rd x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y^2 = k x^2 + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = y \rd y
+      | r = k x \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int y \rd y
+      | r = \int k x \rd x
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y^2} 2
+      | r = \frac {k x^2} 2 + C_1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y^2
+      | r = k x^2 + C
+      | c = reassigning the [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]].
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of First Order ODEs]]
+[[Category:Examples of Separation of Variables]]
+q7mp87fthztpa56s99ft03048efrdar
+\end{proof}<|endoftext|>
+\section{Primitive of Reciprocal of Root of x squared plus k}
+Tags: Primitives involving Root of x squared plus a squared, Primitives involving Root of x squared minus a squared, Primitives involving Root of a squared minus x squared, Primitives involving Reciprocals
+
+\begin{theorem}
+Let $k \in \R$.
+Then:
+:$\displaystyle \int \frac {\d x} {\sqrt {\size {x^2 + k} } } = \map \ln {x + \sqrt {\size {x^2 + k} } } + C$
+\end{theorem}
+
+\begin{proof}
+There are three cases:
+:$0 \le k$
+:$-x^2 < k < 0$
+:$k < -x^2$
+=== $(1): \quad 0 \le k$ ===
+If $0 \le k$ then $k = a^2$ for some $a \in \R$.
+Then [[Primitive of Reciprocal of Root of x squared plus a squared|Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$]] applies:
+:$\displaystyle \int \frac {\d x} {\sqrt {x^2 + a^2} } = \map \ln {x + \sqrt {x^2 + a^2} } + C$
+=== $(2): \quad -x^2 < k < 0$ ===
+If $-x^2 < k < 0$ then:
+:$k = -a^2$ for some $a \in \R$
+and:
+:$x^2 - a^2 > 0$
+Then [[Primitive of Reciprocal of Root of x squared minus a squared|Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$]] applies:
+:$\displaystyle \int \frac {\d x} {\sqrt {x^2 - a^2} } = \map \ln {x + \sqrt {x^2 - a^2} } + C$
+=== $(3): \quad k < -x^2$ ===
+If $k < -x^2$ then:
+:$k = -a^2$ for some $a \in \R$
+and:
+:$a^2 - x^2 > 0$
+Then [[Primitive of Reciprocal of Root of a squared minus x squared/Arcsine Form|Primitive of $\dfrac 1 {\sqrt {a^2 - x^2} }$]] applies:
+:$\displaystyle \int \frac {\d x} {\sqrt {a^2 - x^2} } = \map \ln {x + \sqrt {a^2 - x^2} } + C$
+The result holds for all three cases.
+{{qed}}
+[[Category:Primitives involving Root of x squared plus a squared]]
+[[Category:Primitives involving Root of x squared minus a squared]]
+[[Category:Primitives involving Root of a squared minus x squared]]
+[[Category:Primitives involving Reciprocals]]
+d9jhgj45z2hif4ntthtibtbnusj6dwu
+\end{proof}<|endoftext|>
+\section{Bernoulli's Equation/x^2 dy = (2 x y + y^2) dx}
+Tags: Examples of Bernoulli's Equation
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x^2 \rd y = \paren {2 x y + y^2} \rd x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = - \dfrac {x^2} {x + C}$
+\end{theorem}
+
+\begin{proof}
+Rearranging $(1)$:
+:$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = \dfrac {y^2} {x^2}$
+It can be seen that $(2)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
+where:
+:$\map P x = -\dfrac 2 x$
+:$\map Q x = \dfrac 1 {x^2}$
+:$n = 2$
+and so is an example of [[Definition:Bernoulli's Equation|Bernoulli's equation]].
+By [[Solution to Bernoulli's Equation]] it has the [[Definition:General Solution|general solution]]:
+:$(3): \quad \displaystyle \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$
+where:
+:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
+Thus $\map \mu x$ is evaluated:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - n} \int \map P x \rd x
+      | r = \paren {1 - 2} \int -\dfrac 2 x \rd x
+      | c = 
+}}
+{{eqn | r = 2 \ln x
+      | c = 
+}}
+{{eqn | r = \ln x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \mu x
+      | r = e^{\ln x^2}
+      | c = 
+}}
+{{eqn | r = x^2
+      | c = 
+}}
+{{end-eqn}}
+and so substituting into $(3)$:
+{{begin-eqn}}
+{{eqn | l = x^2 \frac 1 y
+      | r = \paren {1 - 2} \int \dfrac 1 {x^2} x^2 \rd x
+      | c = 
+}}
+{{eqn | r = -\int \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {x^2} y
+      | r = -x + C
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = -\frac {x^2} {x + C}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of Bernoulli's Equation]]
+pkb326kfdtfk18zkhwk69sp4ow2nx18
+\end{proof}<|endoftext|>
+\section{Bernoulli's Equation/2 x y dx + (x^2 + 2 y) dy = 0}
+Tags: Examples of Bernoulli's Equation
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad 2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$
+has the [[Definition:General Solution to Differential Equation|solution]]:
+:$y \paren {x^2 + y} = C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 2 x y \rd x + \paren {x^2 + 2 y} \rd y
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = \dfrac {\d x} {\d y} + \frac 1 {2 y} x
+      | r = -\frac 1 x
+      | c = 
+}}
+{{end-eqn}}
+It can be seen that $(2)$ is in the form:
+:$\dfrac {\d x} {\d y} + \map P y x = \map Q y x^n$
+where:
+:$\map P y = \dfrac 1 {2 y}$
+:$\map Q y = -1$
+:$n = -1$
+and so is an example of [[Definition:Bernoulli's Equation|Bernoulli's equation]].
+By [[Solution to Bernoulli's Equation]] it has the [[Definition:General Solution|general solution]]:
+:$(3): \quad \displaystyle \frac {\map \mu y} {x^{n - 1} } = \paren {1 - n} \int \map Q y \, \map \mu y \rd y + C$
+where:
+:$\map \mu y = e^{\paren {1 - n} \int \map P y \rd y}$
+Thus $\map \mu x$ is evaluated:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - n} \int \map P y \rd y
+      | r = \paren {1 - \paren {-1} } \int \dfrac 1 {2 y} \rd y
+      | c = 
+}}
+{{eqn | r = 2 \frac 1 2 \ln y
+      | c = 
+}}
+{{eqn | r = \ln y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \mu y
+      | r = e^{\ln y}
+      | c = 
+}}
+{{eqn | r = y
+      | c = 
+}}
+{{end-eqn}}
+and so substituting into $(3)$:
+{{begin-eqn}}
+{{eqn | l = \frac y {x^{\paren {-1} - 1} }
+      | r = \paren {1 - \paren {-1} } \int \paren {-1} y \rd y
+      | c = 
+}}
+{{eqn | r = -2 \int y \rd y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y x^2
+      | r = -2 \frac {y^2} 2 + C
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y \paren {x^2 + y}
+      | r = C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of Bernoulli's Equation]]
+0jlmrbe1mo38cnp9og957tw0cmiv2wx
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/y' - (y over x) = k x}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+Let $k \in \R$ be a [[Definition:Real Number|real number]].
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad \dfrac {\d y} {\d x} - \dfrac y x = k x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\dfrac y x = k x + C$
+or:
+:$y = k x^2 + C x$
+\end{theorem}
+
+\begin{proof}
+$(1)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where $\map P x = -\dfrac 1 x$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int -\frac 1 x \rd x
+      | c = 
+}}
+{{eqn | r = -\ln x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = e^{-\ln x}
+      | c = 
+}}
+{{eqn | r = \frac 1 x
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\map {\dfrac {\d} {\d x} } {\dfrac y x} = \dfrac 1 x k x = k$
+and the [[Definition:General Solution|general solution]] is:
+:$\dfrac y x = k x + C$
+or:
+:$y = k x^2 + C x$
+{{qed}}
+[[Category:Examples of Linear First Order ODEs]]
+2c5s81vvm5t4d1tgvkb8zr7atz6asrc
+\end{proof}<|endoftext|>
+\section{First Order ODE/x dy = k y dx}
+Tags: Examples of First Order ODEs, Examples of Separation of Variables, First Order ODE: x dy = k y dx
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad x \rd y = k y \rd x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C x^k$
+\end{theorem}
+
+\begin{proof}
+$(1)$ can be expressed as:
+{{begin-eqn}}
+{{eqn | l = x \rd y
+      | r = k y \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int \dfrac {\d y} y
+      | r = k \int \dfrac {\d x} x
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \ln y
+      | r = k \ln x + \ln C
+      | c = 
+}}
+{{eqn | r = \map \ln {C x^k}
+      | c = 
+}}
+{{eqn | l = y
+      | r = C x^k
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of First Order ODEs]]
+[[Category:Examples of Separation of Variables]]
+[[Category:First Order ODE/x dy = k y dx]]
+i7raos39f81so6kkk5pupb6v1ws4dc6
+\end{proof}<|endoftext|>
+\section{First Order ODE/dy = k y dx}
+Tags: Examples of Linear First Order ODEs, Examples of Separation of Variables, First Order ODE: dy = k y dx
+
+\begin{theorem}
+Let $k \in \R$ be a [[Definition:Real Number|real number]].
+The [[Definition:First Order ODE|first order ODE]]:
+:$\dfrac {\d y} {\d x} = k y$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C e^{k x}$
+\end{theorem}<|endoftext|>
+\section{First Order ODE/(1 over x^3 y^2 + 1 over x) dx + (1 over x^2 y^3 - 1 over y) dy = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y = 0$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = \dfrac 1 {x^3 y^2} + \dfrac 1 x$
+:$\map N {x, y} = \dfrac 1 {x^2 y^3} - \dfrac 1 y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = -\dfrac 2 {x^3 y^3}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = -\dfrac 2 {x^3 y^3}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {\dfrac 1 {x^3 y^2} + \dfrac 1 x} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = -\frac 1 {2 x^2 y^2} + \ln x + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {\dfrac 1 {x^2 y^3} - \dfrac 1 y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = -\frac 1 {2 x^2 y^2} - \ln y + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = -\dfrac 1 {2 x^2 y^2} + \ln x - \ln y$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$-\dfrac 1 {2 x^2 y^2} + \ln x - \ln y = C$
+or:
+:$-\dfrac 1 {2 x^2 y^2} = \ln \dfrac y x + C$
+{{qed}}
+[[Category:Examples of Exact Differential Equation]]
+[[Category:Examples of First Order ODEs]]
+d7d8aehr5sm521jwtav2qh493ooxzmm
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/y' + (y over x) = k x^n}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+Let $k, n \in \R$ be [[Definition:Real Number|real numbers]].
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad \dfrac {\d y} {\d x} + \dfrac y x = k x^n$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \begin{cases}
+\dfrac {k x^{n + 1} } {n + 2} + \dfrac C x & : n \ne -2 \\
+& \\
+\dfrac {k \ln x} x + \dfrac C x & : n = -2
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+$(1)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where:
+:$\map P x = \dfrac 1 x$
+:$\map Q x = k x^n$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int \frac 1 x \rd x
+      | c = 
+}}
+{{eqn | r = \ln x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = e^{\ln x}
+      | c = 
+}}
+{{eqn | r = x
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\map {\dfrac \d {\d x} } {x y} = k x^{n + 1}$
+When $n + 1 \ne -1$, [[Primitive of Power]] is used to obtain:
+:$x y = \dfrac {k x^{n + 2} } {n + 2} + C$
+or:
+:$y = \dfrac {k x^{n + 1} } {n + 2} + \dfrac C x$
+When $n + 1 = -1$, we have:
+:$\map {\dfrac \d {\d x} } {x y} = \dfrac k x$
+and [[Primitive of Reciprocal]] is used, yielding:
+:$x y = k \ln x + C$
+or:
+:$y = \dfrac {k \ln x} x + \dfrac C x$
+{{qed}}
+[[Category:Examples of Linear First Order ODEs]]
+081vts8gx9yr5g78be7x8dw3nbboh71
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/(x^2 + y) dx = x dy}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad \paren {x^2 + y} \rd x = x \rd y$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = x^2 + C x$
+\end{theorem}
+
+\begin{proof}
+Rearranging $(1)$:
+:$(2): \quad \dfrac {\d y} {\d x} - \dfrac y x = x$
+$(2)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where:
+:$\map P x = -\dfrac 1 x$
+:$\map Q x = x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int -\dfrac 1 x \rd x
+      | c = 
+}}
+{{eqn | r = -\ln x
+      | c = 
+}}
+{{eqn | r = \map \ln {\frac 1 x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = \frac 1 x
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\map {\dfrac \d {\d x} } {\dfrac y x} = 1$
+and the [[Definition:General Solution|general solution]] is:
+:$\dfrac y x = x + C$
+or:
+:$y = x^2 + C x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/x y' + y = x^2 cosine x}
+Tags: Examples of Linear First Order ODEs, Linear First Order ODE: x y' + y = x^2 cosine x
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
+\end{theorem}<|endoftext|>
+\section{First Order ODE/(6x + 4y + 3) dx + (3x + 2y + 2) dy = 0}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad \paren {6 x + 4 y + 3} \rd x + \paren {3 x + 2 y + 2} \rd y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$3 x + 2 y + \map \ln {\paren {3 x + 2 y}^2} + x = C$
+\end{theorem}
+
+\begin{proof}
+We put $(1)$ in the form:
+:$\dfrac {\d y} {\d x} = -\dfrac {6 x + 4 y + 3} {3 x + 2 y + 2}$
+Substitute $z = 3 x + 2 y$, which gives:
+:$\dfrac {\d z} {\d x} = 3 + 2 \dfrac {\d y} {\d x}$
+and so:
+:$\dfrac {\d z} {\d x} = -2 \dfrac {2 z + 3} {z + 2} + 3$
+which simplifies down to:
+:$\dfrac {\d z} {\d x} = \dfrac {-z} {z + 2}$
+This is separable:
+:$\displaystyle - \int \frac {z + 2} z \rd z = \int \rd x$
+which gives:
+:$- z - 2 \ln z = x + C$
+which, after the substitutions, gives:
+:$3 x + 2 y + \map \ln {\paren {3 x + 2 y}^2} + x = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/Cosine (x + y) dx = sine (x + y) dx + x sine (x + y) dy}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$\map \cos {x + y} \rd x = \map \sin {x + y} \rd x + x \map \sin {x + y} \rd y$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$x \map \cos {x + y} = C$
+\end{theorem}
+
+\begin{proof}
+Express it in the form:
+:$\paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + x \map \sin {x + y} \rd y$
+Let:
+:$\map M {x, y} = x \map \sin {x + y} - \map \cos {x + y}$
+:$\map N {x, y} = x \map \sin {x + y}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\partial M} {\partial y}
+      | r = x \map \cos {x + y} + \map \sin {x + y}
+      | c = 
+}}
+{{eqn | l = \frac {\partial N} {\partial x}
+      | r = x \map \cos {x + y} + \map \sin {x + y}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {x \map \sin {x + y} - \map \cos {x + y} } \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \map \sin {x + y} - x \map \cos {x + y} - \map \sin {x + y}
+      | c = 
+}}
+{{eqn | r = -x \map \cos {x + y}
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int x \map \sin {x + y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = -x \map \cos {x + y} + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = -x \map \cos {x + y}$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$-x \map \cos {x + y} = C_1$
+or setting $C = -C_1$:
+:$x \map \cos {x + y} = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/x y' + y = f (x)}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad x \, \dfrac {\d y} {\d x} + y = \map f x$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\displaystyle x y = \int \map f x \rd x + C$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = x \dfrac {\d y} {\d x} + y
+      | r = \map f x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \rd y + y \rd x
+      | r = \map f x \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \d {x y}
+      | r = \map f x \rd x
+      | c = [[Product Rule for Differentials]]
+}}
+{{eqn | ll= \leadsto
+      | l = x y
+      | r = \int \map f x \rd x + C
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/x^2 y'' + x y' = 1}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$x^2 y'' + x y' = 1$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \dfrac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]].
+Substitute $p$ for $y'$:
+{{begin-eqn}}
+{{eqn | l = x^2 \dfrac {\d p} {\d x} + x p
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x \dfrac {\d p} {\d x} + p
+      | r = \frac 1 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x p
+      | r = \int \frac {\d x} x
+      | c = [[Linear First Order ODE/x y' + y = f (x)|Linear First Order ODE: $x y' + y = \map f x$]]
+}}
+{{eqn | r = \ln x + C_1
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\d y} {\d x}
+      | r = \frac {\ln x} x + \frac {C_1} x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \int \paren {\frac {\ln x} x + \frac {C_1} x} \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \frac {\paren {\ln x}^2} 2 + C_1 \ln x + C_2
+      | c = [[Primitive of Logarithm of x over x|Primitive of $\dfrac {\ln x} x$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sorgenfrey Line is Lindelöf}
+Tags: Sorgenfrey Line, Lindelöf Spaces
+
+\begin{theorem}
+The [[Definition:Sorgenfrey Line|Sorgenfrey line]] is [[Definition:Lindelöf Space|Lindelöf]].
+\end{theorem}
+
+\begin{proof}
+Let $T = \struct {\R, \tau}$ be the [[Definition:Sorgenfrey Line|Sorgenfrey line]].
+Let $\CC$ be an [[Definition:Open Cover|open cover]] for $\R$.
+Define $\VV = \set {U^\circ_\R: U \in \CC}$
+where $U^\circ_\R$ denotes the [[Definition:Interior (Topology)|interior]] of $U$ in the [[Definition:Real Number Line with Euclidean Topology|real number line $R = \struct {\R, \tau_d}$ with the usual (Euclidean) topology]].
+By definition of [[Definition:Interior (Topology)|interior]]:
+:$\forall U \in \CC: U^\circ_\R \subseteq \R$
+By [[Union is Smallest Superset/Set of Sets|Union is Smallest Superset]]:
+:$W := \bigcup \VV \subseteq \R$ 
+By [[Topological Subspace of Real Number Line is Lindelöf]]:
+:$R_W$ is [[Definition:Lindelöf Space|Lindelöf]]
+where $R_W$ denotes [[Definition:Topological Subspace|topological subspace]] of $R$ on $W$.
+By [[Set is Subset of Union/Set of Sets]]:
+:$\forall A \in \VV: A \subseteq W$
+By [[Intersection with Subset is Subset]]:
+:$\forall A \in \VV: A \cap W = A$
+By definition of [[Definition:Topological Subspace|topological subspace]]:
+:$\forall A \in \VV: A$ is [[Definition:Open Set (Topology)|open]] in $R_W$
+By definition
+:$\VV$ is [[Definition:Open Cover|open cover]]
+By definition of [[Definition:Lindelöf Space|Lindelöf space]]:
+:there exists a [[Definition:Countable Subcover|countable subcover]] $\SS$ of $\VV$
+By definition of $\VV$:
+:$\forall A \in \VV: \exists U \in \CC: A = U^\circ_\R$
+By [[Axiom:Axiom of Choice|Axiom of Choice]] define a [[Definition:Mapping|mapping]] $g: \VV \to \C C$:
+:$\forall A \in \VV: A = \map g A^\circ_\R$
+Define $K = \map {g^\to} \SS$
+where $\map {g^\to} \SS$ denotes the [[Definition:Image of Subset under Mapping|image]] of $\SS$ under $g$.
+Define $Y := \R \setminus \bigcup K$
+By definition of [[Definition:Cover of Set|cover]]:
+:$\R \subseteq \bigcup \CC$
+By definition of [[Definition:Subset|subset]]:
+:$\forall x \in \R: x \in \bigcup \CC$
+By definition of [[Definition:Union of Set of Sets|union]]:
+:$\forall x \in \R: \exists U \in \CC: x \in U$
+By [[Axiom:Axiom of Choice|Axiom of Choice]] define a [[Definition:Mapping|mapping]] $f: \R \to \CC$ such that
+:$\forall x \in \R: x \in \map f x$
+Define $\BB := \set {\hointr x y: x, y \in \R}$
+By definition of the [[Definition:Sorgenfrey Line|Sorgenfrey line]]:
+:$\BB$ is a [[Definition:Analytic Basis|basis]] of $T$.
+By definition of $f$:
+:$\forall x \in \R: \map f x \in \CC$
+By definition of [[Definition:Open Cover|open cover]]:
+:$\forall x \in \R: \map f x$ is [[Definition:Open Set (Topology)|open]]
+By definition of a [[Definition:Analytic Basis|basis]]:
+:$\forall x \in \R: \exists U_x \in \BB: x \in U_x \subseteq \map f x$
+By definition of $\BB$:
+:$\forall x \in \R: \exists y, z \in \R: x \in \hointr y z \subseteq \map f x$
+By definition of [[Definition:Half-Open Real Interval|half-open real interval]]:
+:$\forall x \in \R: \exists z \in \R: x \in \hointr x z \subseteq \map f x$
+By [[Axiom:Axiom of Choice|Axiom of Choice]] define a [[Definition:Mapping|mapping]] $k: \R \to \BB$:
+:$\forall x \in \R: \exists z \in \R: x \in \map k x = \hointr x z \subseteq \map f x$
+We will prove that
+:$(1): \quad \forall x, y \in Y: x \ne y \implies \paren x \cap \map k y = \O$
+{{Mistake|For $\paren x$ should it really be $\map k x$?}}
+Let $x, y \in Y$ such that $x \ne y$.
+{{AimForCont}}:
+:$\paren x \cap \map k y \ne \O$
+By definitions of [[Definition:Empty Set|empty set]] and [[Definition:Set Intersection|intersection]]:
+:$\exists s: s \in \map k x \land s \in \map k y$
+By definition of $k$:
+:$\exists z_1 \in \R: \map k x = \hointr x {z_1} \subseteq \map f x$
+and
+:$\exists z_2 \in \R: \map k y = \hointr y {z_2}$
+By [[Trichotomy Law for Real Numbers]]:
+:$x < y$ or $x > y$
+{{WLOG}}, suppose $x < y$
+By definition of [[Definition:Half-Open Real Interval|half-open real interval]]:
+:$x \le s < z_1$ and $y \le s < z_2$
+Then:
+$y < z_1$
+By definition of [[Definition:Open Real Interval|open real interval]]:
+:$y \in \openint x {z_1}$
+By [[Open Real Interval is Open Set]]:
+:$\openint x {z_1}$ is [[Definition:Open Set (Topology)|topologically open]] in $R$
+By definition of [[Definition:Subset|subset]]:
+:$\openint x {z_1} \subseteq \hointr x {z_1}$
+By [[Subset Relation is Transitive]]:
+:$\openint x {z_1} \subseteq \map f x$
+By [[Interior of Subset]]:
+:$\openint x {z_1}^\circ_\R \subseteq \map f x^\circ_\R$
+By [[Interior of Open Set]]:
+:$\openint x {z_1}^\circ_\R \openint x z_1$
+{{mistake|The relation is missing in the above -- subset or equals? Research needed}}
+By definition of [[Definition:Subset|subset]]:
+:$y \in \map f x^\circ_\R$
+By definition of $\VV$:
+:$\map f x^\circ_\R \in \VV$
+By definition of [[Definition:Union of Set of Sets|union]]:
+$y \in W$
+By definition of [[Definition:Interior (Topology)|interior]]:
+:$\forall A \in \SS: A \subseteq \map g A$
+By [[Set Union Preserves Subsets]]:
+:$W \subseteq \bigcup \SS \subseteq \bigcup K$
+By definition of [[Definition:Subset|subset]]:
+:$y \in \bigcup K$
+This [[Definition:Contradiction|contradicts]] $y \in Y$ by definition of [[Definition:Set Difference|difference]].
+Thus:
+:$\map k x \cap \map k y = \O$
+By [[Set of Pairwise Disjoint Intervals is Countable]]
+:$\map {k^\to} Y$ is [[Definition:Countable Set|countable]]
+We will prove that
+:$k \restriction_Y$ is an [[Definition:Injection|injection]]
+Let $x, y \in Y$ such that
+:$\map {k \restriction_Y} x = \map {k \restriction_Y} Y$
+{{AimForCont}}:
+:$x \ne y$
+Then by $(1)$:
+:$\map k x \cap \map k y = \O$
+By definition of [[Definition:Restriction of Mapping|restriction of mapping]]:
+:$\map {k \restriction_Y} x = \map k x$
+and
+:$\map {k \restriction_Y} y = \map k y$
+By definition of $k$:
+:$x \in \map k x$
+So:
+:$x \in \map k y$
+This contradicts:
+:$\map k x \cap k \map k y = \O$
+Thus $x = y$.
+By [[Injection to Image is Bijection]]:
+:$k \restriction_Y: Y \to \map {k^\to} Y$ is a [[Definition:Bijection|bijection]]
+By definitions of [[Definition:Set Equivalence|set equivalence]] and [[Definition:Cardinality|cardinality]]:
+:$\card Y = \card {\map {k^\to} Y}$
+where $\card Y$ denotes the [[Definition:Cardinality|cardinality]] of $Y$.
+By [[Cardinality of Image of Set not greater than Cardinality of Set]]:
+:$\card K \le \card \SS$ and $\card {\map {f^\to} Y} \le \card Y$
+By [[Countable iff Cardinality not greater than Aleph Zero]]:
+:$\card \SS \le \aleph_0$ and $\card Y \le \aleph_0$
+Then:
+:$\card K \le \aleph_0$ and $\card {\map {f^\to} Y} \le \aleph_0$
+By [[Countable iff Cardinality not greater than Aleph Zero]]:
+:$K$ is [[Definition:Countable Set|countable]] and $\map {f^\to} Y$ is [[Definition:Countable Set|countable]]
+Thus by [[Countable Union of Countable Sets is Countable]]:
+:$\GG := K \cup \map {f^\to} Y$ is [[Definition:Countable Set|countable]]
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$K \subseteq \CC$ and $\map {f^\to} Y \subseteq \CC$
+thus by [[Set Union Preserves Subsets/Corollary|corollary of Set Union Preserves Subsets]]:
+:$\GG \subseteq \CC$
+It remains to prove that
+:$\GG$ is [[Definition:Cover of Set|cover]] for $\R$
+Let $x \in \R$.
+By [[Union Distributes over Union/Sets of Sets|Union Distributes over Union: Sets of Sets]]:
+:$\bigcup \GG = \paren {\bigcup K} \cup \bigcup \map {f^\to} Y$
+{{AimForCont}}:
+:$ x \notin \bigcup \GG$
+By definition of [[Definition:Set Union|union]]:
+:$x \notin \bigcup K$ and $x \notin \bigcup \map {f^\to} Y$
+By definition of [[Definition:Set Difference|difference]]:
+:$x in Y$
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\map f x \in \map {f^\to} Y$
+By definition of $f$:
+:$x \in \map f x$
+By definition of [[Definition:Union of Set of Sets|union]]:
+:$x \in \bigcup \map {f^\to} Y$
+This contradicts $x \notin \bigcup \map {f^\to} Y$.
+Thus the result by [[Proof by Contradiction]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(y^2 exp x y + cosine x) dx + (exp x y + x y exp x y) dy = 0}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \paren {y^2 e^{x y} + \cos x} \rd x + \paren {e^{x y} + x y e^{x y} } \rd y = 0$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$y e^{x y} + \sin x = C$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\map M {x, y} = y^2 e^{x y} + \cos x$
+:$\map N {x, y} = e^{x y} + x y e^{x y}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = 2 y e^{x y} + x y^2 e^{x y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = y e^{x y} + x y^2 e^{x y} + y e^{x y}
+      | c = 
+}}
+{{eqn | r = 2 y e^{x y} + x y^2 e^{x y}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {y^2 e^{x y} + \cos x} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = y e^{x y} + \sin x + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {e^{x y} + x y e^{x y} } \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \frac {e^{x y} } x + y e^{x y} - \frac {e^{x y} } x + \map h x
+      | c = [[Primitive of x by Exponential of a x|Primitive of $x e^{a x}$]]
+}}
+{{eqn | r = y e^{x y} + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = y e^{x y} + \sin x$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$y e^{x y} + \sin x = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/y' ln (x - y) = 1 + ln (x - y)}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \dfrac {\d y} {\d x} \map \ln {x - y} \d x = 1 + \map \ln {x - y}$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$\paren {x - y} \map \ln {x - y} = C - y$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be expressed as:
+:$\paren {1 + \map \ln {x - y} } \rd x - \map \ln {x - y} \rd y = 0$
+Let:
+:$\map M {x, y} = 1 + \map \ln {x - y}$
+:$\map N {x, y} = -\map \ln {x - y}$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = -\frac 1 {x - y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = -\frac 1 {x - y}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \left({1 + \map \ln {x - y} }\right) \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = x + \paren {x - y} \map \ln {x - y} - \paren {x - y} + \map g y
+      | c = [[Primitive of Logarithm of x|Primitive of $\ln x$]]
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \left({-\map \ln {x - y} }\right) \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \paren {x - y} \map \ln {x - y} - \paren {x - y} + \map h x
+      | c = [[Primitive of Logarithm of x|Primitive of $\ln x$]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l = \map f {x, y}
+      | r = x + \paren {x - y} \map \ln {x - y} - \paren {x - y}
+      | c = 
+}}
+{{eqn | r = \paren {x - y} \map \ln {x - y} + y
+      | c = 
+}}
+{{end-eqn}}
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+{{begin-eqn}}
+{{eqn | l = \paren {x - y} \map \ln {x - y} + y
+      | r = C
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {x - y} \map \ln {x - y}
+      | r = C - y
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/y' + 2 x y = exp -x^2}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$\dfrac {\d y} {\d x} + 2 x y = \map \exp {-x^2}$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \paren {x + C} \map \exp {-x^2}$
+\end{theorem}
+
+\begin{proof}
+$(1)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where:
+:$\map P x = 2 x$
+:$\map Q x = \map \exp {-x^2}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = \int 2 x \rd x
+      | c = 
+}}
+{{eqn | r = x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = \map \exp {x^2}
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+:$\map {\dfrac {\d} {\d x} } {\map \exp {x^2} y} = \map \exp {-x^2} \map \exp {x^2} = 1$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \map \exp {x^2} y
+      | r = \int 1 \rd x
+      | c = 
+}}
+{{eqn | r = x + C
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \paren {x + C} \map \exp {-x^2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(y^2 - 3 x y - 2 x^2) dx = (x^2 - x y) dy}
+Tags: Examples of Homogeneous Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \paren {y^2 - 3 x y - 2 x^2} \rd x = \paren {x^2 - x y} \rd y$
+is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]] with [[Definition:General Solution|solution]]:
+:$y^2 x^2 - 2 y x^3 + x^4 = C$
+\end{theorem}
+
+\begin{proof}
+$(1)$ can also be rendered:
+:$\dfrac {\rd y} {\rd x} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$
+Let:
+:$\map M {x, y} = y^2 - 3 x y - 2 x^2$
+:$\map N {x, y} = x^2 - x y$
+Put $t x, t y$ for $x, y$:
+{{begin-eqn}}
+{{eqn | l = \map M {t x, t y}
+      | r = \paren {t y}^2 - 3 t x t y - 2 \paren {t x}^2
+      | c = 
+}}
+{{eqn | r = t^2 \paren {y^2 - 3 x y - 2 x^2}
+      | c = 
+}}
+{{eqn | r = t^2 \, \map M {x, y}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \map N {t x, t y}
+      | r = \paren {t x}^2 - t x t y
+      | c = 
+}}
+{{eqn | r = t^2 \paren {x^2 - x y}
+      | c = 
+}}
+{{eqn | r = t^2 \, \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Thus both $M$ and $N$ are [[Definition:Homogeneous Real Function|homogeneous functions]] of [[Definition:Degree of Homogeneous Real Function|degree]] $2$.
+Thus, by definition, $(1)$ is a [[Definition:Homogeneous Differential Equation|homogeneous differential equation]].
+By [[Solution to Homogeneous Differential Equation]], its solution is:
+:$\displaystyle \ln x = \int \frac {\d z} {\map f {1, z} - z} + C$
+where:
+:$\map f {x, y} = \dfrac {y^2 - 3 x y - 2 x^2} {x^2 - x y}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \ln x
+      | r = \int \frac {\d z} {\frac {z^2 - 3 z - 2} {1 - z} - z} + C_1
+      | c = 
+}}
+{{eqn | r = \int \frac {1 - z} {2 z^2 - 4 z - 2} \rd z + C_1
+      | c = 
+}}
+{{eqn | r = -\frac 1 4 \int \frac {4 z - 4} {2 z^2 - 4 z - 2} \rd z + C_1
+      | c = 
+}}
+{{eqn | r = -\frac 1 4 \ln \map \{2 z^2 - 4 z - 2} + C_1
+      | c = [[Primitive of Function under its Derivative]]
+}}
+{{end-eqn}}
+Substituting back for $z$ and tidying up, the result is obtained:
+:$y^2 x^2 - 2 y x^3 + x^4 = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/(1 + x^2) y' + 2 x y = 4 x^3}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$\paren {1 + x^2} \dfrac {\d y} {\d x} + 2 x y = 4 x^3$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \dfrac {x^4} {1 + x^2} + \dfrac C {1 + x^2}$
+\end{theorem}
+
+\begin{proof}
+It is noticed that:
+:$\dfrac {\d} {\d x} \paren {1 + x^2} = 2 x$
+and so $(1)$ can be rendered in the form:
+{{begin-eqn}}
+{{eqn | l = \map \d {\paren {1 + x^2} y}
+      | r = 4 x^3
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {1 + x^2} y
+      | r = \int 4 x^3 \rd x
+      | c = 
+}}
+{{eqn | r = x^4 + C
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \frac {x^4} {1 + x^2} + \frac C {1 + x^2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/exp x sine y dx + exp x cos y dy = y sine x y dx + x sine x y dy}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad e^x \sin y \rd x + e^x \cos y \rd y = y \sin x y \rd x + x \sin x y \rd y$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$e^x \sin y + \cos x y = C$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be expressed as:
+:$\paren {e^x \sin y - y \sin x y} \rd x + \paren {e^x \cos y - x \sin x y} \rd y = 0$
+Let:
+:$\map M {x, y} = e^x \sin y - y \sin x y$
+:$\map N {x, y} = e^x \cos y - x \sin x y$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = e^x \cos y - x y \cos x y - y \sin x y
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = e^x \cos y - x y \cos x y - y \sin x y
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {e^x \sin y - y \sin x y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = e^x \sin y + \cos x y + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {e^x \cos y - x \sin x y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = e^x \sin y + \cos x y + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = e^x \sin y + \cos x y$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$e^x \sin y + \cos x y = C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(1 + x^2) y' + x y = 0}
+Tags: Examples of Linear First Order ODEs, Examples of Separation of Variables
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$\paren {1 + x^2} y' + x y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \dfrac C {\sqrt {1 + x^2} }$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \paren {1 + x^2} \dfrac {\d y} {\d x}
+      | r = -x y
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \int \frac {\d y} y
+      | r = -\int \frac {x \rd x} {1 + x^2}
+      | c = [[Separation of Variables]]
+}}
+{{eqn | ll= \leadsto
+      | l = \ln y
+      | r = -\frac 1 2 \, \map \ln {1 + x^2} + C
+      | c = [[Primitive of x over x squared plus a squared|Primitive of $\dfrac x {x^2 + a^2}$]]
+}}
+{{eqn | ll= \leadsto
+      | l = \ln y
+      | r = \ln \frac {C_1} {\sqrt {1 + x^2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \frac C {\sqrt {1 + x^2} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Examples of Linear First Order ODEs]]
+[[Category:Examples of Separation of Variables]]
+q1ev8n414r0vr7rxjt0ng5cia3utj6i
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/(1 + x^2) y'' + x y' = 0}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$\paren {1 + x^2} y'' + x y' = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2$
+\end{theorem}
+
+\begin{proof}
+The proof proceeds by using [[Solution of Second Order Differential Equation with Missing Dependent Variable]].
+Substitute $p$ for $y'$:
+{{begin-eqn}}
+{{eqn | l = \paren {1 + x^2} \dfrac {\d p} {\d x} + x p
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p = \dfrac {\d y} {\d x}
+      | r = \frac {C_1} {\sqrt {1 + x^2} }
+      | c = [[First Order ODE/(1 + x^2) y' + x y = 0|First Order ODE: $\paren {1 + x^2} y' + x y = 0$]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = C_1 \int \frac 1 {\sqrt {1 + x^2} } \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2
+      | c = [[Primitive of Reciprocal of Root of x squared plus a squared|Primitive of $\dfrac 1 {\sqrt{x^2 + a^2} }$]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/(x exp y + y - x^2) dy = (2 x y - exp y - x) dx}
+Tags: Examples of Exact Differential Equation, Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad \paren {x e^y + y - x^2} \rd y = \paren {2 x y - e^y - x} \rd x$
+is an [[Definition:Exact Differential Equation|exact differential equation]] with [[Definition:General Solution|solution]]:
+:$2 x e^y + x^2 + y^2 - 2 x^2 y = C$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be expressed in the form:
+:$\paren {e^y + x - 2 x y} \rd x + \paren {x e^y + y - x^2} rd y = 0$
+Let:
+:$\map M {x, y} = e^y + x - 2 x y$
+:$\map N {x, y} = x e^y + y - x^2$
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial M} {\partial y}
+      | r = e^y - 2 x
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial N} {\partial x}
+      | r = e^y - 2 x
+      | c = 
+}}
+{{end-eqn}}
+Thus $\dfrac {\partial M} {\partial y} = \dfrac {\partial N} {\partial x}$ and $(1)$ is seen by definition to be [[Definition:Exact Differential Equation|exact]].
+By [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$\map f {x, y} = C$
+where:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\partial f} {\partial x}
+      | r = \map M {x, y}
+      | c = 
+}}
+{{eqn | l = \dfrac {\partial f} {\partial y}
+      | r = \map N {x, y}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \map M {x, y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = \int \paren {e^y + x - 2 x y} \rd x + \map g y
+      | c = 
+}}
+{{eqn | r = x e^y + \frac {x^2} 2 - x^2 y + \map g y
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = f
+      | r = \int \map N {x, y} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = \int \paren {x e^y + y - x^2} \rd y + \map h x
+      | c = 
+}}
+{{eqn | r = x e^y + \frac {y^2} 2 - x^2 y + \map h x
+      | c = 
+}}
+{{end-eqn}}
+Thus:
+:$\map f {x, y} = x e^y + \dfrac {x^2 + y^2} 2 - x^2 y$
+and by [[Solution to Exact Differential Equation]], the [[Definition:General Solution|solution]] to $(1)$ is:
+:$2 x e^y + x^2 + y^2 - 2 x^2 y = C$
+after multiplying by $2$ to clear the fraction.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{First Order ODE/exp x (1 + x) dx = (x exp x - y exp y) dy}
+Tags: Examples of First Order ODEs
+
+\begin{theorem}
+The [[Definition:First Order Ordinary Differential Equation|first order ordinary differential equation]]:
+:$(1): \quad e^x \paren {1 + x} \rd x = \paren {x e^x - y e^y} \rd y$
+has the [[Definition:General Solution|solution]]:
+:$2 x e^x e^{-y} + y^2 + C$
+\end{theorem}
+
+\begin{proof}
+Dividing $(1)$ through by $e^y$:
+:$(2): \quad \paren {1 + x} e^x e^{-y} \rd x = \paren {x e^x e^{-y} - y} \rd y$
+Let $z = x e^x e^{-y}$.
+Then:
+:$\dfrac {\d z} {\d x} = \paren {1 + x} e^x e^{-y} - x e^x e^{-y} \dfrac {\d y} {\d x}$
+or:
+:$\d z = \paren {1 + x} e^x e^{-y} \rd x - x e^x e^{-y} \rd y$
+Substituting $z$ and $\mathrm d z$ in $(2)$:
+:$\d z = \paren {z - y} \rd y - z \rd y = - y \rd y$
+which directly leads to:
+:$\displaystyle \int \rd z = -\int y \rd y$
+from which:
+:$z = - \dfrac {y^2} 2 + k$
+Substituting for $z$ and letting $C = 2 k$:
+:$2 x e^x e^{-y} + y^2 + C$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Two Stage Radioactive Decay}
+Tags: Decay Equation
+
+\begin{theorem}
+Let $R_A$ be a [[Definition:Radioactive Isotope|radioactive isotope]] which [[Definition:Radioactive Decay|decays]] into another [[Definition:Radioactive Isotope|radioactive isotope]] $R_B$ with the [[Definition:Rate Constant|rate constant]] $k_1$.
+Let $R_B$ [[Definition:Radioactive Decay|decay]] into a third [[Definition:Chemical Element|element]] $R_C$ with the [[Definition:Rate Constant|rate constant]] $k_2$.
+Let the initial quantity of $R_A$ be $x_0$.
+Let the amounts of $R_A$ and $R_B$ present at [[Definition:Time|time]] $t$ be $x$ and $y$ respectively.
+Then:
+:$y = \begin{cases}
+\dfrac {k_1 x_0} {k_2 - k_1} \left({e^{-k_1 t} - e^{-k_2 t} }\right) & : k_1 \ne k_2 \\
+& \\
+k_1 x_0 t e^{-k_1 t} & : k_1 = k_2
+\end{cases}$
+\end{theorem}<|endoftext|>
+\section{General Solution of Riccati Equation from Particular Solution}
+Tags: Riccati Equation
+
+\begin{theorem}
+Consider the [[Definition:Riccati Equation|Riccati equation]]:
+:$(1): \quad y' = \map p x + \map q x y + \map r x y^2$
+Let $\map {y_1} x$ be a [[Definition:Particular Solution|particular solution]] to $(1)$.
+Then the [[Definition:General Solution|general solution]] to $(1)$ has the form:
+:$\map y x = \map {y_1} x + \map z x$
+where $\map z x$ is the [[Definition:General Solution|general solution]] to the [[Definition:Bernoulli's Equation|Bernoulli equation]]:
+:$z' - \paren {q - 2 r y_1} z = r z^2$
+\end{theorem}
+
+\begin{proof}
+Let $\map y x = \map {y_1} x + \map z x$ be a [[Definition:Particular Solution|particular solution]] to $(1)$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d x}
+      | r = y_1' + z'
+      | c = 
+}}
+{{eqn | r = \map p x + \map q x \paren {y_1 + z} + \map r x \paren {y_1 + z}^2
+      | c = 
+}}
+{{eqn | r = \map p x + \map q x \paren {y_1 + z} + \map r x \paren {y_1^2 + 2 y_1 z + z^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z'
+      | r = \map q x z + \map r x \paren {2 y_1 z + z^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = z' - \paren {\map q x + 2 y_1 \map r x z}
+      | r = \map r x z^2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Riccati Equation/y' = (y over x) + x^3 y^2 - x^5}
+Tags: Examples of Riccati Equation
+
+\begin{theorem}
+The [[Definition:Riccati Equation|Riccati equation]]:
+:$(1): \quad y' = \dfrac y x + x^3 y^2 - x^5$
+has the [[Definition:General Solution|general solution]]:
+:$C \exp \dfrac {2 x^5} 5 = \dfrac {y - x} {y + x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen by inspection that:
+:$\map {y_1} x = x$
+is a [[Definition:Particular Solution|particular solution]] to $(1)$.
+Thus from [[General Solution of Riccati Equation from Particular Solution]]:
+:$y = x + \map z x$
+where:
+:$z' - \paren {\dfrac 1 x + 2 x^4} z = x^3 z^2$
+From [[Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2|Bernoulli's Equation: $y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$]]:
+:$z = \dfrac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}$
+Thus:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = x + \frac {2 x} {C \map \exp {-\dfrac {2 x^5} 5} - 1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {y - x} \paren {C \map \exp {-\dfrac {2 x^5} 5} - 1}
+      | r = 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y C \map \exp {-\dfrac {2 x^5} 5} - y - x C \map \exp {-\dfrac {2 x^5} 5} + x
+      | r = 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y C \map \exp {-\dfrac {2 x^5} 5} - y - x C \map \exp {-\dfrac {2 x^5} 5} - x
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {y - x} C \map \exp {-\dfrac {2 x^5} 5}
+      | r = y + x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y - x} {y + x}
+      | r = C \map \exp {\dfrac {2 x^5} 5}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bernoulli's Equation/y' - (1 over x + 2 x^4) y = x^3 y^2}
+Tags: Examples of Bernoulli's Equation
+
+\begin{theorem}
+The [[Definition:First Order ODE|first order ODE]]:
+:$(1): \quad y' - \paren {\dfrac 1 x + 2 x^4} y = x^3 y^2$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = \dfrac {2 x} {C \, \map \exp {-\dfrac {2 x^5} 5} - 1}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x y^n$
+where:
+:$\map P x = -\paren {\dfrac 1 x + 2 x^4}$
+:$\map Q x = x^3$
+:$n = 2$
+and so is an example of [[Definition:Bernoulli's Equation|Bernoulli's equation]].
+By [[Solution to Bernoulli's Equation]] it has the [[Definition:General Solution|general solution]]:
+:$(3): \quad \displaystyle \frac {\map \mu x} {y^{n - 1} } = \paren {1 - n} \int \map Q x \, \map \mu x \rd x + C$
+where:
+:$\map \mu x = e^{\paren {1 - n} \int \map P x \rd x}$
+Thus $\map \mu x$ is evaluated:
+{{begin-eqn}}
+{{eqn | l = \paren {1 - n} \int \map P x \rd x
+      | r = \paren {1 - 2} \int -\paren {\dfrac 1 x + 2 x^4} \rd x
+      | c = 
+}}
+{{eqn | r = \ln x + \dfrac {2 x^5} 5
+      | c = 
+}}
+{{eqn | r = \map \ln {x \exp \dfrac {2 x^5} 5}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map \mu x
+      | r = e^{\map \ln {x \exp \frac {2 x^5} 5} }
+      | c = 
+}}
+{{eqn | r = x \exp \dfrac {2 x^5} 5
+      | c = 
+}}
+{{end-eqn}}
+and so substituting into $(3)$:
+{{begin-eqn}}
+{{eqn | l = \frac x y \exp \dfrac {2 x^5} 5
+      | r = \paren {1 - 2} \int x^3 x \exp \dfrac {2 x^5} 5 \rd x
+      | c = 
+}}
+{{eqn | r = -\int x^4 \exp \dfrac {2 x^5} 5 \rd x
+      | c = 
+}}
+{{eqn | r = -\frac {\exp \dfrac {2 x^5} 5} 2 + \frac C 2
+      | c = by using the substitution $w = x^5$
+}}
+{{eqn | ll= \leadsto
+      | l = \frac x y
+      | r = \frac C 2 \map \exp {-\dfrac {2 x^5} 5} - \frac 1 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 1 y
+      | r = \frac C {2 x} \map \exp {-\dfrac {2 x^5} 5} - \frac 1 {2 x}
+      | c = 
+}}
+{{eqn | r = \frac {C \, \map \exp {-\dfrac {2 x^5} 5} - 1} {2 x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = \frac {2 x} {C \, \map \exp {-\dfrac {2 x^5} 5} - 1}
+      | c = 
+}}
+{{end-eqn}}
+Hence the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$ is:
+:$y = \dfrac {2 x} {C \, \map \exp {-\dfrac {2 x^5} 5} - 1}$
+{{qed}}
+[[Category:Examples of Bernoulli's Equation]]
+276qfq1i7tb9y18ikpfs6e09myqzajq
+\end{proof}<|endoftext|>
+\section{Motion of Body with Variable Mass}
+Tags: Dynamics
+
+\begin{theorem}
+Let $B$ be a [[Definition:Body|body]] undergoing a [[Definition:Force|force]] $\mathbf F$.
+Let $B$ be travelling at a [[Definition:Velocity|velocity]] $\mathbf v$ at [[Definition:Time|time]] $t$.
+Let [[Definition:Mass|mass]] travelling at a [[Definition:Velocity|velocity]] $\mathbf v + \mathbf w$ be added to $B$ at a [[Definition:Rate|rate]] of $\dfrac {\d m} {\d t}$.
+Let $m$ be the [[Definition:Mass|mass]] of $B$ at [[Definition:Time|time]] $t$.
+Then the equation of motion of $B$ is given by:
+:$\mathbf w \dfrac {\d m} {\d t} + \mathbf F = m \dfrac {\d \mathbf v} {\d t}$
+\end{theorem}
+
+\begin{proof}
+From [[Newton's Second Law of Motion]]:
+:$\mathbf F = \map {\dfrac \d {\d t} } {m \mathbf v}$
+Then the added [[Definition:Momentum|momentum]] being added to $B$ by the [[Definition:Mass|mass]] being added to it is given by:
+:$\paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \paren {\mathbf v + \mathbf w} \dfrac {\d m} {\d t} + \mathbf F
+      | r = \map {\dfrac \d {\d t} } {m \mathbf v}
+      | c = 
+}}
+{{eqn | r = m \dfrac {\d \mathbf v} {\d t} + \mathbf v \dfrac {\d m} {\d t}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \mathbf w \dfrac {\d m} {\d t} + \mathbf F
+      | r = m \dfrac {\d \mathbf v} {\d t}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Burnout Velocity of Upward Rocket under Constant Gravity}
+Tags: Rocket Science
+
+\begin{theorem}
+Let $R$ be a [[Definition:Rocket|rocket]] whose structural [[Definition:Mass|mass]] is $m_1$.
+Let $R$ contain fuel of initial [[Definition:Mass|mass]] $m_2$.
+Let $R$ be fired straight up from the surface of a [[Definition:Planet|planet]] whose [[Definition:Gravitational Field|gravitational field]] exerts an [[Acceleration Due to Gravity]] of $g$, assumed constant.
+Let $R$ burn fuel at a constant [[Definition:Rate|rate]] $a$, producing a constant [[Definition:Exhaust Velocity|exhaust velocity]] $b$ relative to $R$.
+Let all [[Definition:Force|forces]] on $R$ except that due to the [[Definition:Gravitational Field|gravitational field]] be neglected.
+Then the [[Definition:Burnout Velocity|burnout velocity]] of $R$ is given by:
+:$v_b = b \ln \left({1 + \dfrac {m_1} {m_2} }\right) - \dfrac {g m_2} a$
+\end{theorem}<|endoftext|>
+\section{Burnout Height of Upward Rocket under Constant Gravity}
+Tags: Rocket Science
+
+\begin{theorem}
+Let $R$ be a [[Definition:Rocket|rocket]] whose structural [[Definition:Mass|mass]] is $m_1$.
+Let $R$ contain fuel of initial [[Definition:Mass|mass]] $m_2$.
+Let $R$ be fired straight up from the surface of a [[Definition:Planet|planet]] whose [[Definition:Gravitational Field|gravitational field]] exerts an [[Acceleration Due to Gravity]] of $g$, assumed constant.
+Let $R$ burn fuel at a constant [[Definition:Rate|rate]] $a$, with a constant [[Definition:Exhaust Velocity|exhaust velocity]] $b$ relative to $R$.
+Let all forces on $R$ except that due to  the [[Definition:Gravitational Field|gravitational field]] be neglected.
+Then the [[Definition:Burnout Height|burnout height]] of $R$ is given by:
+:$h_b = -\dfrac {g m_2^2} {2 a^2} + \dfrac {b m_2} a + \dfrac {b m_1} a \ln \dfrac {m_1} {m_1 + m_2}$
+\end{theorem}<|endoftext|>
+\section{Linear First Order ODE/y' = x + y}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad \dfrac {\d y} {\d x} = x + y$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C e^x - x - 1$
+\end{theorem}
+
+\begin{proof}
+Rearranging $(1)$:
+:$(2): \quad \dfrac {\d y} {\d x} - y = x$
+$(2)$ is a [[Definition:Linear First Order ODE|linear first order ODE]] in the form:
+:$\dfrac {\d y} {\d x} + \map P x y = \map Q x$
+where:
+:$\map P x = -1$
+:$\map Q x = x$
+Thus:
+{{begin-eqn}}
+{{eqn | l = \int \map P x \rd x
+      | r = -\int \rd x
+      | c = 
+}}
+{{eqn | r = - x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{\int P \rd x}
+      | r = e^{-x}
+      | c = 
+}}
+{{eqn | r = \frac 1 {e^x}
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Solution by Integrating Factor]], $(1)$ can be rewritten as:
+{{begin-eqn}}
+{{eqn | l = \map {\dfrac {\d} {\d x} } {\dfrac y {e^x} }
+      | r = \frac x {e^x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac y {e^x}
+      | r = \int \frac x {e^x} \rd x
+      | c = 
+}}
+{{eqn | r = -\frac 1 {e^x} \paren {x + 1} + C
+      | c = [[Primitive of x by Exponential of a x|Primitive of $x e^{a x}$]]
+}}
+{{eqn | ll= \leadsto
+      | l = y
+      | r = C e^x - x - 1
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear First Order ODE/y' = x + y/y(0) = 1}
+Tags: Examples of Linear First Order ODEs
+
+\begin{theorem}
+The [[Definition:Linear First Order ODE|linear first order ODE]]:
+:$(1): \quad \dfrac {\d y} {\d x} = x + y$
+with [[Definition:Initial Condition|initial condition]]:
+:$\map y 0 = 1$
+has the [[Definition:Particular Solution to Differential Equation|particular solution]]:
+:$y = 2 e^x - x - 1$
+\end{theorem}
+
+\begin{proof}
+From [[Linear First Order ODE/y' = x + y|Linear First Order ODE: $y' = x + y$]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C e^x - x - 1$
+Setting $y = 1$ when $x = 0$ gives:
+:$1 = C + 1$
+from which $C = 2$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions}
+Tags: Linear Second Order ODEs
+
+\begin{theorem}
+Let $\map P x$, $\map Q x$ and $\map R x$ be [[Definition:Continuous Real Function|continuous real functions]] on a [[Definition:Closed Real Interval|closed real interval]] $\closedint a b$.
+Let $x_0$ be any point in $\closedint a b$.
+Let $y_0$ and ${y_0}'$ be [[Definition:Real Number|real numbers]].
+Then the [[Definition:Linear Second Order Ordinary Differential Equation|linear second order ordinary differential equation]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = \map R x$
+has a [[Definition:Unique|unique]] [[Definition:Particular Solution to Differential Equation|particular solution]] $\map y x$ on $\closedint a b$ such that:
+:$\map y {x_0} = y_0$
+and:
+:$\map {y'} {x_0} = {y_0}'$
+\end{theorem}
+
+\begin{proof}
+Let $z = \dfrac {\d y} {\d x}$.
+Then a [[Definition:Solution to Differential Equation|solution]] to $(1)$ will yield a [[Definition:Particular Solution to Differential Equation|particular solution]] to:
+:$(2): \quad \begin{cases}
+\dfrac {\d y} {\d x} = z & , \map y {x_0} = y_0 \\
+& \\
+\dfrac {\d z} {\d x} = -\map P x \dfrac {\d y} {\d x} - \map Q x y + \map R x & , \map z {x_0} = {y_0}'
+\end{cases}$
+From [[Lipschitz Condition on Linear ODE of Continuous Functions]], $(2)$ satisfies the [[Definition:Lipschitz Condition|Lipschitz condition]].
+Hence [[Picard's Existence Theorem]] applies.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution}
+Tags: Linear Second Order ODEs
+
+\begin{theorem}
+Consider the [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = \map R x$
+Let $\map {y_g} x$ be the [[Definition:General Solution to Differential Equation|general solution]] of the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(2): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+Let $\map {y_p} x$ be a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(1)$.
+Then $\map {y_g} x + \map {y_p} x$ is the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$.
+\end{theorem}
+
+\begin{proof}
+Let $\map {y_g} {x, C_1, C_2}$ be a [[Definition:General Solution to Differential Equation|general solution]] of $(2)$.
+Note that $C_1$ and $C_2$ are the two [[Definition:Arbitrary Constant|arbitrary constants]] that are to be expected of a [[Definition:Second Order ODE|second order ODE]].
+Let $\map {y_p} x$ be a certain fixed [[Definition:Particular Solution to Differential Equation|particular solution]] of $(1)$.
+Let $\map y x$ be an arbitrary [[Definition:Particular Solution to Differential Equation|particular solution]] of $(1)$.
+Then:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \paren {y - y_p}'' + \map P x \paren {y - y_p}' + \map Q x \paren {y - y_p}
+      | c = 
+}}
+{{eqn | r = \paren {y'' + \map P x y' + \map Q x y} - \paren {y_p'' + \map P x y_p' + \map Q x y_p}
+      | c = 
+}}
+{{eqn | r = \map R x - \map R x
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+We have that $\map {y_g} {x, C_1, C_2}$ is a [[Definition:General Solution to Differential Equation|general solution]] of $(2)$.
+Thus:
+:$\map y x - \map {y_p} x = \map {y_g} {x, C_1, C_2}$
+or:
+:$\map y x = \map {y_g} {x, C_1, C_2} + \map {y_p} x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Trivial Solution of Homogeneous Linear 2nd Order ODE}
+Tags: Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$\dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+has the [[Definition:Particular Solution|particular solution]]:
+:$\map y x = 0$
+that is, the [[Definition:Constant Mapping|zero constant function]].
+This [[Definition:Particular Solution|particular solution]] is referred to as the '''trivial solution'''.
+\end{theorem}
+
+\begin{proof}
+We have:
+:$\map {\dfrac {\d} {\d x} } 0 = 0$
+and so:
+:$\map {\dfrac {\d^2} {\d x^2} } 0 = 0$
+from which:
+:$\dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+Hence the result.
+{{Qed}}
+\end{proof}<|endoftext|>
+\section{Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE}
+Tags: Homogeneous LSOODEs
+
+\begin{theorem}
+Let $c_1$ and $c_2$ be [[Definition:Real Number|real numbers]].
+Let $\map {y_1} x$ and $\map {y_2} x$ be [[Definition:Particular Solution|particular solutions]] to the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+Then:
+:$c_1 \, \map {y_1} x + c_2 \, \map {y_2} x$
+is also a [[Definition:Particular Solution|particular solution]] to $(1)$.
+That is, a [[Definition:Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE|linear combination]] of [[Definition:Particular Solution|particular solutions]] to a [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]] is also a [[Definition:Particular Solution|particular solution]] to that [[Definition:Homogeneous Linear Second Order ODE|ODE]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | o = 
+      | r = \paren {c_1 \, \map {y_1} x + c_2 \, \map {y_2} x}'' + \map P x \paren {c_1 \, \map {y_1} x + c_2 \, \map {y_2} x}' + \map Q x \paren {c_1 \, \map {y_1} x + c_2 \, \map {y_2} x}
+      | c = 
+}}
+{{eqn | r = \paren {c_1 \, \map {y_1''} x + c_2 \, \map {y_2''} x} + \map P x \paren {c_1 \, \map {y_1'} x + c_2 \, \map {y_2'} x} + \map Q x \paren {c_1 \, \map {y_1} x + c_2 \, \map {y_2} x}
+      | c = [[Linear Combination of Derivatives]]
+}}
+{{eqn | r = c_1 \paren {\map {y_1''} x + \map P x \, \map {y_1'} x + \map Q x \, \map {y_1} x} + c_2 \paren {\map {y_2''} x + \map P x \, \map {y_2'} x + \map Q x \, \map {y_2} x}
+      | c = 
+}}
+{{eqn | r = c_1 \cdot 0 + c_2 \cdot 0
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Solution of Linear 2nd Order ODE Tangent to X-Axis}
+Tags: Linear Second Order ODEs
+
+\begin{theorem}
+Let $\map {y_p} x$ be a [[Definition:Particular Solution|particular solution]] to the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let there exist $\xi \in \closedint a b$ such that the [[Definition:Curve|curve]] in the [[Definition:Cartesian Plane|cartesian plane]] described by $y = \map {y_p} x$ is [[Definition:Tangent to Curve|tangent]] to the [[Definition:X-Axis|$x$-axis]] at $\xi$.
+Then $\map {y_p} x$ is the [[Definition:Constant Mapping|zero constant function]]:
+:$\forall x \in \closedint a b: \map {y_p} x = 0$
+\end{theorem}
+
+\begin{proof}
+{{AimForCont}} $y_p$ is not the [[Definition:Constant Mapping|zero constant function]].
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]], there exists another [[Definition:Particular Solution|particular solution]] to $(1)$ such that $y_1$ and $y_2$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+At the point $\xi$:
+:$\map {y_p} \xi = 0$
+:$\map { {y_p}'} \xi = 0$
+Taking the [[Definition:Wronskian|Wronskian]] of $y_p$ and $y_2$:
+:$\map W {y_p, y_2} = y_p {y_2}' - {y_p}' y_2$
+But at $\xi$ this works out as [[Definition:Zero (Number)|zero]].
+It follows from [[Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent]] that $y_p$ and $y_2$ cannot be [[Definition:Linearly Independent Real Functions|linearly independent]] after all.
+From this contradiction, $y_p$ must the [[Definition:Constant Mapping|zero constant function]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Solutions of Linear 2nd Order ODE have Common Zero iff Linearly Dependent}
+Tags: Linear Second Order ODEs
+
+\begin{theorem}
+Let $\map {y_1} x$ and $\map {y_2} x$ be [[Definition:Particular Solution|particular solutions]] to the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $y_1$ and $y_2$ both have a [[Definition:Zero of Function|zero]] for the same value of $x$ in $\closedint a b$.
+Then $y_1$ and $y_2$ are [[Definition:Constant|constant]] [[Definition:Multiple|multiples]] of each other.
+That is, $y_1$ and $y_2$ are [[Definition:Linearly Dependent Real Functions|linearly dependent]].
+\end{theorem}
+
+\begin{proof}
+Let $\xi \in \closedint a b$ be such that $\map {y_1} \xi = \map {y_2} \xi = 0$.
+Consider the [[Definition:Wronskian|Wronskian]] $\map W {y_1, y_2}$ at $\xi$:
+{{begin-eqn}}
+{{eqn | l = \map W {\map {y_1} \xi, \map {y_2} \xi}
+      | r = \map {y_1} \xi \map { {y_2}'} \xi - \map {y_2} \xi \map { {y_1}'} \xi
+      | c = 
+}}
+{{eqn | r = 0 \cdot \map { {y_2}'} \xi - 0 \cdot \map { {y_1}'} \xi
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+From [[Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE]]:
+:$\forall x \in \closedint a b: \map W {\map {y_1} \xi, \map {y_2} \xi} = 0$
+and so from [[Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent]]:
+:$y_1$ and $y_2$ are [[Definition:Linearly Dependent Real Functions|linearly dependent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Function is Linearly Dependent with Zero Function}
+Tags: Real Analysis
+
+\begin{theorem}
+Let $f \left({x}\right)$ be a [[Definition:Real Function|real function]] defined on a [[Definition:Closed Real Interval|closed interval]] $\left[{a \,.\,.\, b}\right]$.
+Let $g \left({x}\right)$ be the [[Definition:Constant Mapping|constant zero function]] on $\left[{a \,.\,.\, b}\right]$:
+:$\forall x \in \left[{a \,.\,.\, b}\right]: g \left({x}\right) = 0$
+Then $f$ and $g$ are [[Definition:Linearly Dependent Real Functions|linearly dependent]] on $\left[{a \,.\,.\, b}\right]$.
+\end{theorem}
+
+\begin{proof}
+We have that:
+:$\forall x \in \left[{a \,.\,.\, b}\right]: g \left({x}\right) = 0 = 0 \times f \left({x}\right)$
+and $0 \in \R$.
+Hence the result by definition of [[Definition:Linearly Dependent Real Functions|linearly dependent real functions]].
+\end{proof}<|endoftext|>
+\section{Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution}
+Tags: Homogeneous LSOODEs
+
+\begin{theorem}
+Let $\map {y_1} x$ and $\map {y_2} x$ be [[Definition:Particular Solution|particular solutions]] to the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $y_1$ and $y_2$ be [[Definition:Linearly Independent Real Functions|linearly independent]].
+Then the [[Definition:General Solution|general solution]] to $(1)$ is:
+:$y = C_1 \map {y_1} x + C_2 \map {y_2} x$
+where $C_1 \in \R$ and $C_2 \in \R$ are [[Definition:Arbitrary Constant|arbitrary constants]].
+\end{theorem}
+
+\begin{proof}
+Let $\map y x$ be any [[Definition:Particular Solution|particular solution]] to $(1)$ on $\closedint a b$.
+It is to be shown that [[Definition:Constant|constants]] $C_1$ and $C_2$ can be found such that:
+:$\map y x = C_1 \map {y_1} x + C_2 \map {y_2} x$
+for all $x \in \closedint a b$.
+By [[Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions]]:
+:a [[Definition:Particular Solution|particular solution]] to $(1)$ over $\closedint a b$ is completely determined by:
+::its [[Definition:Value of Element under Mapping|value]]
+:and:
+::the [[Definition:Value of Element under Mapping|value]] of its [[Definition:Derivative|derivative]]
+at a single point.
+From [[Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE]]:
+:$C_1 \map {y_1} x + C_2 \map {y_2} x$
+is a [[Definition:Particular Solution|particular solution]] to $(1)$ over $\closedint a b$
+We also have:
+:$\map y x$
+is a [[Definition:Particular Solution|particular solution]] to $(1)$ over $\closedint a b$
+Thus it is sufficient to prove that:
+:$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that:
+::$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$
+:and:
+::$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$
+For this system to be solvable for $C_1$ and $C_2$ it is necessary that:
+:$\begin{vmatrix}
+\map {y_1} x & \map {y_2} x \\
+\map { {y_1}'} x & \map { {y_2}'} x \\
+\end{vmatrix} = \map {y_1} x \map { {y_2}'} x - \map {y_2} x \map { {y_1}'} x \ne 0$
+That is, that the [[Definition:Wronskian|Wronskian]] $\map W {y_1, y_2} \ne 0$ at $x_0$.
+From [[Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE]]:
+:if $\map W {y_1, y_2} \ne 0$ at $x_0$, then $\map W {y_1, y_2} \ne 0$ for all $x \in \closedint a b$.
+Hence it does not matter what point is taken for $x_0$; if the [[Definition:Wronskian|Wronskian]] is non-zero at one such point, it will be non-zero for all such points.
+From [[Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent]]:
+:$W \left({y_1, y_2}\right) = 0$ for all $x \in \closedint a b$ {{iff}} $y_1$ and $y_2$ are [[Definition:Linearly Dependent Real Functions|linearly dependent]].
+But we have that $y_1$ and $y_2$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+Hence:
+:$\forall x \in \closedint a b: \map W {y_1, y_2} \ne 0$
+and so:
+:$\exists x_0 \in \closedint a b: \exists C_1, C_2 \in \R$ such that:
+::$ C_1 \map {y_1} {x_0} + C_2 \map {y_2} {x_0} = \map y {x_0}$
+:and:
+::$ C_1 \map { {y_1}'} {x_0} + C_2 \map { {y_2}'} {x_0} = \map y {x_0}$
+The result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE}
+Tags: Homogeneous LSOODEs
+
+\begin{theorem}
+Let $\map {y_1} x$ and $\map {y_2} x$ be [[Definition:Particular Solution|particular solutions]] to the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Let $y_1$ and $y_2$ be [[Definition:Linearly Independent Real Functions|linearly independent]].
+Then their [[Definition:Wronskian|Wronskian]] is either never zero, or zero everywhere on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {W'} {y_1, y_2}
+      | r = \paren {y_1 {y_2}'' + {y_1}' {y_2}'} - \paren {y_2 {y_1}'' + {y_2}' {y_1}'}
+      | c = [[Product Rule]]
+}}
+{{eqn | r = y_1 {y_2}'' - y_2 {y_1}''
+      | c = 
+}}
+{{end-eqn}}
+Because $y_1$ and $y_2$ are both [[Definition:Particular Solution|particular solutions]] of $(1)$:
+{{begin-eqn}}
+{{eqn | n = 2
+      | l = {y_1}'' + \map P x {y_1}' + \map Q x y_1
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 3
+      | l = {y_2}'' + \map P x {y_2}' + \map Q x y_2
+      | r = 0
+      | c = 
+}}
+{{eqn | n = 4
+      | ll= \leadsto
+      | l = y_2 {y_1}'' + \map P x y_2 {y_1}' + \map Q x y_2 y_1
+      | r = 0
+      | c = $(2)$ multiplied by $y_2$
+}}
+{{eqn | n = 5
+      | l = y_1 {y_2}'' + \map P x y_1 {y_2}' + \map Q x y_1 y_2
+      | r = 0
+      | c = $(3)$ multiplied by $y_1$
+}}
+{{eqn | n = 6
+      | ll= \leadsto
+      | l = \paren {y_1 {y_2}'' - y_2 {y_1}''} + \map P x \paren {y_1 {y_2}' - y_2 {y_1}'}
+      | r = 0
+      | c = $(5)$ subtracted from $(6)$
+}}
+{{end-eqn}}
+That is:
+:$\dfrac {\d P} {\d W} + P W = 0$
+This is a [[Definition:Linear First Order Ordinary Differential Equation|linear first order ODE]].
+From [[Solution to Linear First Order Ordinary Differential Equation]]:
+:$W = C e^{-\int P \rd x}$
+The [[Definition:Real Exponential Function|exponential function]] is never zero:
+Therefore:
+:$W = 0 \iff C = 0$
+and the result follows.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Real Number Line is Lindelöf}
+Tags: Lindelöf Spaces, Real Number Line with Euclidean Topology
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Then $\struct {\R, \tau_d}$ is [[Definition:Lindelöf Space|Lindelöf]].
+\end{theorem}
+
+\begin{proof}
+From [[Real Number Line is Second-Countable]] we have that $\struct {\R, \tau_d}$ is a [[Definition:Second-Countable Space|second-countable space]].
+The result follows from [[Second-Countable Space is Lindelöf]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent}
+Tags: Homogeneous LSOODEs
+
+\begin{theorem}
+Let $\map {y_1} x$ and $\map {y_2} x$ be [[Definition:Particular Solution|particular solutions]] to the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+on a [[Definition:Closed Real Interval|closed interval]] $\closedint a b$.
+Then:
+:$y_1$ and $y_2$ are [[Definition:Linearly Dependent Real Functions|linearly dependent]]
+{{iff}}:
+:the [[Definition:Wronskian|Wronskian]] $\map W {y_1, y_2}$ of $y_1$ and $y_2$ is zero everywhere on $\closedint a b$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $y_1$ and $y_2$ are [[Definition:Linearly Dependent Real Functions|linearly dependent]].
+Suppose either $y_1$ or $y_2$ is zero everywhere on $\closedint a b$.
+Then either ${y_1}'$ or ${y_2}'$ is also zero everywhere on $\closedint a b$.
+Thus:
+:$y_1 {y_2}' - y_2 {y_1}' = 0$
+and so by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.
+Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$.
+{{WLOG}}, it follows by definition of [[Definition:Linearly Dependent Real Functions|linearly dependent]] that:
+:$y_2 = C y_1$
+for some $C \in \R$.
+Thus by [[Derivative of Constant Multiple]]:
+:${y_2}' = C {y_1}'$
+Hence:
+{{begin-eqn}}
+{{eqn | l = C
+      | r = \frac {y_2} {y_1}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_2}'
+      | r = \frac {y_2} {y_1} {y_1}'
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y_1 {y_2}'
+      | r = y_2 {y_1}'
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y_1 {y_2}' - y_2 {y_1}'
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Hence by definition $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $\map W {y_1, y_2} = 0$ everywhere on $\closedint a b$.
+One way for this to happen is for either $y_1$ or $y_2$ to be zero everywhere on $\closedint a b$.
+{{WLOG}}, suppose $y_1$ is zero everywhere on $\closedint a b$.
+Then by [[Real Function is Linearly Dependent with Zero Function]], $y_1$ and $y_2$ are [[Definition:Linearly Dependent Real Functions|linearly dependent]].
+Suppose that neither $y_1$ nor $y_2$ is zero everywhere on $\closedint a b$.
+As $y_1$ is [[Definition:Continuous Real Function|continuous]], there is some [[Definition:Closed Real Interval|closed interval]]:
+:$\closedint c d \subseteq \closedint a b$
+on which:
+:$\forall x \in \closedint c d: \map {y_1} x \ne 0$
+Since $\map W {y_1, y_2} = 0$:
+:$\dfrac {y_1 {y_2}' - y_2 {y_1}'} { {y_1}^2} = 0$
+on $\closedint c d$.
+By [[Quotient Rule for Derivatives]], this is the same as:
+:$\paren {\dfrac {y_2} {y_1} }' = 0$
+[[Definition:Integration|Integrating]] {{WRT|Integration}} $x$, this gives:
+:$\dfrac {y_2} {y_1} = k$
+Explicitly:
+:$\forall x \in \closedint c d: \map {y_2} x = k \, \map {y_1} x$
+As $\map {y_2} x$ and $k \, \map {y_1} x$ have equal values on $\closedint c d$, their [[Definition:Derivative|derivatives]] are equal there as well.
+From [[Existence and Uniqueness of Solution for Linear Second Order ODE with two Initial Conditions]], it follows that:
+:$\forall x \in \closedint a b: \map {y_2} x = k \, \map {y_1} x$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs, Linear Second Order ODE: y'' + y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 \sin x + C_2 \cos x$
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/y'' + y = 0/y(0) = 2, y'(0) = 3}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + y = 0$
+with [[Definition:Initial Condition|initial conditions]]:
+:$\map y 0 = 2$
+:$\map {y'} 0 = 3$
+has the [[Definition:Particular Solution of Differential Equation|particular solution]]:
+:$y = 3 \sin x + 2 \cos x$
+\end{theorem}
+
+\begin{proof}
+From [[Linear Second Order ODE/y'' + y = 0|Linear Second Order ODE: $y'' + y = 0$]], the [[Definition:General Solution|general solution]] of $(1)$ is:
+:$y = C_1 \sin x + C_2 \cos x$
+[[Definition:Differentiation|Differentiating]] {{WRT|Differentiation}} $x$:
+:$y' = C_1 \cos x - C_2 \sin x$
+Thus for the [[Definition:Initial Condition|initial conditions]]:
+{{begin-eqn}}
+{{eqn | l = \map y 0
+      | r = C_1 \sin 0 + C_2 \cos 0
+      | c = 
+}}
+{{eqn | r = 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_1 \times 0 + C_2 \times 1
+      | r = 2
+      | c = [[Sine of Zero is Zero]], [[Cosine of Zero is One]]
+}}
+{{eqn | ll= \leadsto
+      | l = C_2
+      | r = 2
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map {y'} 0
+      | r = C_1 \cos 0 - C_2 \sin 0
+      | c = 
+}}
+{{eqn | r = 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_1 \times 1 - C_2 \times 0
+      | r = 3
+      | c = [[Sine of Zero is Zero]], [[Cosine of Zero is One]]
+}}
+{{eqn | ll= \leadsto
+      | l = C_1
+      | r = 3
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linearly Independent Solutions of y'' - y = 0}
+Tags: Linear Second Order ODE: y'' - y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - y = 0$
+has [[Definition:Solution to Differential Equation|solutions]]:
+:$y_1 = e^x$
+:$y_2 = e^{-x}$
+which are [[Definition:Linearly Independent Real Functions|linearly independent]].
+\end{theorem}
+
+\begin{proof}
+We have that:
+{{begin-eqn}}
+{{eqn | l = \frac {\d} {\d x} \, e^x
+      | r = e^x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | l = \frac {\d} {\d x} \, e^{-x}
+      | r = -e^{-x}
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d^2} {\d x^2} \, e^x
+      | r = e^x
+      | c = 
+}}
+{{eqn | l = \frac {\d^2} {\d x^2} \, e^{-x}
+      | r = e^{-x}
+      | c =  
+}}
+{{end-eqn}}
+Hence it can be seen by inspection that:
+{{begin-eqn}}
+{{eqn | l = \map {y_1} x
+      | r = e^x
+}}
+{{eqn | l = \map {y_2} x
+      | r = e^{-x}
+}}
+{{end-eqn}}
+are [[Definition:Solution to Differential Equation|solutions]] to $(1)$.
+Calculating the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$:
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = \begin{vmatrix} e^x & e^{-x} \\ e^x & -e^{-x} \end{vmatrix}
+      | c = 
+}}
+{{eqn | r = -e^x e^{-x} - e^x e^{-x}
+      | c = 
+}}
+{{eqn | r = -2
+      | c = 
+}}
+{{end-eqn}}
+So the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$ is never zero.
+Thus from [[Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent]]:
+:$y_1$ and $y_2$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0}
+Tags: Examples of Cauchy-Euler Equation, Linear Second Order ODE: x^2 y'' - 2 x y' + 2 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x^2 y'' - 2 x y' + 2 y = 0$
+has the [[Definition:General Solution|general solution]]:
+:$y = C_1 x + C_2 x^2$
+on any [[Definition:Closed Real Interval|closed real interval]] which does not contain $0$.
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0/y(1) = 3, y'(1) = 5}
+Tags: Linear Second Order ODE: x^2 y'' - 2 x y' + 2 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x^2 y'' - 2 x y' + 2 y = 0$
+with [[Definition:Initial Condition|initial conditions]]:
+:$\map y 1 = 3$
+:$\map {y'} 1 = 5$
+has the [[Definition:Particular Solution of Differential Equation|particular solution]]:
+:$y = x + 2 x^2$
+\end{theorem}
+
+\begin{proof}
+From [[Linear Second Order ODE/x^2 y'' - 2 x y' + 2 y = 0|Linear Second Order ODE: $x^2 y'' - 2 x y' + 2 y = 0$]], the [[Definition:General Solution|general solution]] of $(1)$ is:
+:$y = C_1 x + C_2 x^2$
+[[Definition:Differentiation|Differentiating]] {{WRT|Differentiation}} $x$:
+:$y' = C_1 + 2 C_2 x$
+Thus for the [[Definition:Initial Condition|initial conditions]]:
+{{begin-eqn}}
+{{eqn | l = \map y 1
+      | r = C_1 \times 1 + C_2 \times 1^2
+      | c = 
+}}
+{{eqn | r = 3
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = C_1 + C_2
+      | r = 3
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map {y'} 1
+      | r = C_1 + 2 C_2 \times 1
+      | c = 
+}}
+{{eqn | r = 5
+      | c = 
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = C_1 + 2 C_2
+      | r = 5
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_2
+      | r = 2
+      | c = $(3) - (2)$
+}}
+{{eqn | ll= \leadsto
+      | l = C_1 + 2
+      | r = 3
+      | c = substituting for $C_2$ in $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = C_1
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 3 y' + 2 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs, Linear Second Order ODE: y'' - 3 y' + 2 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 3 y' + 2 y = 0$
+has the [[Definition:General Solution|general solution]]:
+:$y = C_1 e^x + C_2 e^{2 x}$
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 3 y' + 2 y = 0/y(0) = -1, y'(0) = 1}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 3 y' + 2 y = 0$
+with [[Definition:Initial Condition|initial conditions]]:
+:$\map y 0 = -1$
+:$\map {y'} 0 = 1$
+has the [[Definition:Particular Solution of Differential Equation|particular solution]]:
+:$y = -3 e^x + 2 e^{2 x}$
+\end{theorem}
+
+\begin{proof}
+From [[Linear Second Order ODE/y'' - 3 y' + 2 y = 0|Linear Second Order ODE: $y'' - 3 y' + 2 y = 0$]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^x + C_2 e^{2 x}$
+[[Definition:Differentiation|Differentiating]] {{WRT|Differentiation}} $x$:
+:$y' = C_1 e^x + 2 C_2 e^{2 x}$
+Thus for the [[Definition:Initial Condition|initial conditions]]:
+{{begin-eqn}}
+{{eqn | l = \map y 0
+      | r = C_1 e^0 + C_2 e^{2 \times 0}
+      | c = 
+}}
+{{eqn | r = -1
+      | c = 
+}}
+{{eqn | n = 2
+      | ll= \leadsto
+      | l = C_1 + C_2
+      | r = -1
+      | c = 
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \map {y'} 1
+      | r = C_1 e^0 + 2 C_2 e^{2 \times 0}
+      | c = 
+}}
+{{eqn | r = 1
+      | c = 
+}}
+{{eqn | n = 3
+      | ll= \leadsto
+      | l = C_1 + 2 C_2
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_2
+      | r = 2
+      | c = $(3) - (2)$
+}}
+{{eqn | ll= \leadsto
+      | l = C_1 + 2
+      | r = -1
+      | c = substituting for $C_2$ in $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = C_1
+      | r = -3
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 4 y' + 4 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs, Linear Second Order ODE: y'' - 4 y' + 4 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 4 y' + 4 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{2 x} + C_2 x e^{2 x}$
+\end{theorem}<|endoftext|>
+\section{Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another}
+Tags: Homogeneous LSOODEs
+
+\begin{theorem}
+Let $\map {y_1} x$ be a [[Definition:Particular Solution|particular solution]] to the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+such that $y_1$ is not the [[Trivial Solution of Homogeneous Linear 2nd Order ODE|trivial solution]].
+Then there exists a standard procedure to determine another [[Definition:Particular Solution|particular solution]] $\map {y_2} x$ of $(1)$ such that $y_1$ and $y_2$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+\end{theorem}
+
+\begin{proof}
+Let $\map {y_1} x$ be a [[Trivial Solution of Homogeneous Linear 2nd Order ODE|non-trivial]] [[Definition:Particular Solution|particular solution]] to $(1)$.
+Thus, for all $C \in \R$, $C y_1$ is also a [[Trivial Solution of Homogeneous Linear 2nd Order ODE|non-trivial]] [[Definition:Particular Solution|particular solution]] to $(1)$.
+Let $\map v x$ be a [[Definition:Real Function|function]] of $x$ such that:
+:$(2): \quad \map {y_2} x = \map v x \, \map {y_1} x$
+is a [[Definition:Particular Solution|particular solution]] to $(1)$ such that $y_1$ and $y_2$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+Thus:
+:$(3): \quad {y_2}'' + \map P x {y_2}' + \map Q x y = 0$
+[[Definition:Differentiation|Differentiating]] $(2)$ twice {{WRT|Differentiation}} $x$:
+:${y_2}' = \map {v'} x \, \map {y_1} x + \map v x \, \map { {y_1}'} x$
+and:
+:${y_2}'' = \map {v''} x \, \map {y_1} x + 2 \, \map {v'} x \, \map { {y_1}'} x + \map v x \, \map { {y_1}''} x$
+Substituting these into $(3)$ and rearranging:
+:$v paren { {y_1}'' + P {y_1}' + Q y_1} + v'' y_1 + v' \paren {2 {y_1}'' + P y_1} = 0$
+Since $y_1$ is a [[Definition:Particular Solution|particular solution]] of $(1)$, this reduces to:
+:$v'' y_1 + v' \paren {2 {y_1}'' + P y_1} = 0$
+which can be expressed as:
+:$\dfrac {v''} {v'} = -2 \dfrac { {y_1}'} {y_1} - P$
+[[Definition:Integration|Integration]] gives:
+:$\displaystyle \ln v' = - 2 \ln y_1 - \int P \rd x$
+This leads to:
+:$v' = \dfrac 1 { {y_1}^2} e^{-\int P \rd x}$
+and so:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+From [[Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution]], $y_1$ and $v y_1$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Induced Solution to Homogeneous Linear Second Order ODE is Linearly Independent with Inducing Solution}
+Tags: Homogeneous LSOODEs
+
+\begin{theorem}
+Let $\map {y_1} x$ be a [[Definition:Particular Solution|particular solution]] to the [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + \map P x \dfrac {\d y} {\d x} + \map Q x y = 0$
+such that $y_1$ is not the [[Trivial Solution of Homogeneous Linear 2nd Order ODE|trivial solution]].
+Let $\map {y_2} x$ be the [[Definition:Real Function|real function]] defined as:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+Then $y_2$ and $y_1$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+\end{theorem}
+
+\begin{proof}
+This will be demonstrated by calculating the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$ and demonstrating that it is non-zero everywhere.
+First we take the [[Definition:Derivative|derivative]] of $v$ {{WRT}} $x$:
+:$v' = \dfrac 1 { {y_1}^2} e^{- \int P \rd x}$
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = {{Defof|Wronskian}}
+}}
+{{eqn | r = y_1 \paren {v y_1}' - v y_1 {y_1}'
+      | c = Definition of $y_2$
+}}
+{{eqn | r = y_1 \paren {v' y_1 + v {y_1}'} - v y_1 {y_1}'
+      | c = [[Product Rule]]
+}}
+{{eqn | r = {y_1}^2 v' 
+      | c = simplifying
+}}
+{{eqn | r = {y_1}^2 \dfrac 1 { {y_1}^2} e^{-\int P \rd x}
+      | c = substituting for $v'$
+}}
+{{eqn | r = e^{-\int P \rd x}
+      | c = simplifying
+}}
+{{end-eqn}}
+As $\displaystyle -\int P \rd x$ is a [[Definition:Real Function|real function]], $e^{-\int P \rd x}$ is non-zero wherever it is defined.
+Hence from [[Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent]], $y_1$ and $y_2$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/x^2 y'' + x y' - y = 0}
+Tags: Examples of Cauchy-Euler Equation, Linear Second Order ODE: x^2 y'' + x y' - y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x^2 y'' + x y' - y = 0$
+has the [[Definition:General Solution|general solution]]:
+:$y = C_1 x + \dfrac {C_2} x$
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/y'' - y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs, Linear Second Order ODE: y'' - y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 e^x + C_2 e^{-x}$
+\end{theorem}<|endoftext|>
+\section{Second Order ODE/x y'' + 3 y' = 0}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x y'' + 3 y' = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 + \dfrac {C_2} {x^2}$
+\end{theorem}
+
+\begin{proof}
+Note that:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y'
+      | r = 0
+      | c = [[Derivative of Constant]]
+}}
+{{eqn | ll= \leadsto
+      | l = y''
+      | r = 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+and so by inspection:
+:$y_1 = 1$
+is a [[Definition:Particular Solution|particular solution]] of $(1)$.
+$(1)$ can be expressed as:
+:$(2): \quad y'' + \dfrac 3 x y' = 0$
+which is in the form:
+:$y'' + \map P x y' + \map Q x y = 0$
+where:
+:$\map P x = \dfrac 3 x$
+:$\map Q x = 0$
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int \dfrac 3 x \rd x
+      | c = 
+}}
+{{eqn | r = 3 \ln x
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | r = \ln x^3
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{-\ln x^3}
+      | c = 
+}}
+{{eqn | r = \frac 1 {x^3}
+      | c =  
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \frac 1 {x^3} \rd x
+      | c = 
+}}
+{{eqn | r = -\frac 1 {2 x^2}
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = -\frac 1 {2 x^2}
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$y = C_1 + k \paren {-\frac 1 {2 x^2} }$
+where $k$ is arbitrary.
+Setting $C_2 = - \dfrac k 2$ yields the result:
+:$y = C_1 + \dfrac {C_2} {x^2}$
+{{qed}}
+</onlyinclude>
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/x^2 y'' + x y' - 4 y = 0}
+Tags: Examples of Cauchy-Euler Equation, Linear Second Order ODE: x^2 y'' + x y' - 4 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x^2 y'' + x y' - 4 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 x^2 + \dfrac {C_2} {x^2}$
+\end{theorem}<|endoftext|>
+\section{Legendre's Differential Equation/(1 - x^2) y'' - 2 x y' + 2 y = 0}
+Tags: Examples of Legendre's Differential Equation
+
+\begin{theorem}
+The special case of [[Definition:Legendre's Differential Equation|Legendre's differential equation]]:
+:$(1): \quad \paren {1 - x^2} y'' - 2 x y' + 2 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 x + C_2 \paren {\dfrac x 2 \, \map \ln {\dfrac {1 + x} {1 - x} } - 1}$
+\end{theorem}
+
+\begin{proof}
+Note that:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}'
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}''
+      | r = 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+and so by inspection:
+:$y_1 = x$
+is a [[Definition:Particular Solution|particular solution]] of $(1)$.
+$(1)$ can be expressed as:
+:$(2): \quad y'' - \dfrac {2 x} {1 - x^2} y' + \dfrac 2 {1 - x^2} y = 0$
+which is in the form:
+:$y'' + \map P x y' + \map Q x y = 0$
+where:
+:$\map P x = - \dfrac {2 x} {1 - x^2}$
+:$\map Q x = \dfrac 2 {1 - x^2}$
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int \paren {-\dfrac {2 x} {1 - x^2} } \rd x
+      | c = 
+}}
+{{eqn | r = \map \ln {1 - x^2}
+      | c = [[Primitive of Function under its Derivative]]
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{-\map \ln {1 - x^2} }
+      | c = 
+}}
+{{eqn | r = \frac 1 {1 - x^2}
+      | c =  
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \dfrac 1 {x^2} \frac 1 {1 - x^2} \rd x
+      | c = 
+}}
+{{eqn | r = \int \dfrac 1 {x^2 \paren {1 - x^2} } \rd x
+      | c = 
+}}
+{{eqn | r = -\frac 1 x + \frac 1 2 \map \ln {\frac {1 + x} {1 - x} }
+      | c = [[Primitive of Reciprocal of x squared by a squared minus x squared|Primitive of $\dfrac 1 {x^2 \paren {a^2 - x^2} }$]]
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = \paren {-\frac 1 x + \frac 1 2 \map \ln {\frac {1 + x} {1 - x} } } x
+      | c = 
+}}
+{{eqn | r = -1 + \frac x 2 \map \ln {\frac {1 + x} {1 - x} }
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$y = C_1 x + C_2 \paren {\dfrac x 2 \, \map \ln {\dfrac {1 + x} {1 - x} } - 1}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Topological Subspace of Real Number Line is Lindelöf}
+Tags: Lindelöf Spaces, Real Number Line with Euclidean Topology
+
+\begin{theorem}
+Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
+Let $W$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $\R$.
+Then $R_W$ is [[Definition:Lindelöf Space|Lindelöf]]
+where $R_W$ denotes the [[Definition:Topological Subspace|topological subspace]] of $R$ on $W$.
+\end{theorem}
+
+\begin{proof}
+Let $\CC$ be a [[Definition:Open Cover|open cover]] for $W$.
+Define $Q := \set {\openint a b: a, b \in \Q}$
+Define a [[Definition:Mapping|mapping]] $h: Q \to \Q \times \Q$:
+:$\forall \openint a b \in Q: \map h {\openint a b} = \tuple {a, b}$
+It is easy to see by definition that
+:$h$ is an [[Definition:Injection|injection]].
+By [[Injection iff Cardinal Inequality]]:
+:$\card Q \le \card {\Q \times \Q}$
+where $\card Q$ deontes the [[Definition:Cardinality|cardinality]] of $Q$.
+By [[Rational Numbers are Countably Infinite]]:
+:$\Q$ is [[Definition:Countably Infinite Set|countably infinite]].
+By definition of [[Definition:Countably Infinite Set|countably infinite]]:
+:there exists a [[Definition:Bijection|bijection]] $\Q \to \N$
+By definitions of [[Definition:Set Equality|set equality]] and [[Definition:Cardinality|cardinality]]:
+:$\card \Q = \card \N$
+By [[Aleph Zero equals Cardinality of Naturals]]:
+:$\card \Q = \aleph_0$
+By [[Cardinal Product Equal to Maximum]]:
+:$\card {\Q \times \Q} = \max \set {\aleph_0, \aleph_0} = \aleph_0$
+By [[Countable iff Cardinality not greater than Aleph Zero]]:
+:$Q$ is [[Definition:Countable Set|countable]].
+By definition of [[Definition:Cover of Set|cover]]:
+:$W \subseteq \bigcup \CC$
+By definition of [[Definition:Union of Set of Sets|imion]]:
+:$\forall x \in W: \exists U \in \CC: x \in U$
+By [[Axiom:Axiom of Choice|Axiom of Choice]] define a [[Definition:Mapping|mapping]] $f: W \to \CC$:
+:$\forall x \in W: x \in \map f x$
+We will prove that
+:$\forall x \in W: \exists A \in Q: x \in A \land A \cap W \subseteq \map f x$
+Let $x \in W$.
+By definition of [[Definition:Open Cover|open cover]]:
+:$\map f x$ is [[Definition:Open Set (Topology)|open]] in $R_W$.
+By definition of [[Definition:Topological Subspace|topological subspace]]:
+:there exists U a [[Definition:Subset|subset]] of $\R$ such that
+::$U$ is [[Definition:Open Set (Topology)|open]] in $R$ and $U \cap W = \map f x$
+By definition of $f$:
+:$x \in \map f x$
+By definition of [[Definition:Open Set of Metric Space|open set in metric space]]:
+:$\exists r > 0: \map {B_r} x \subseteq U$
+:$\openint {x - r} {x + r} \subseteq U$
+By [[Between two Real Numbers exists Rational Number]]:
+:$\exists q \in \Q: x - r < q < x$ and $\exists p \in \Q: x < p < x + r$
+By definition of $Q$:
+:$\openint q p \in Q$
+Thus by definition of [[Definition:Open Real Interval|open real interval]]:
+:$x \in \openint q p \subseteq \openint {x - r} {x + r}$
+By [[Subset Relation is Transitive]]:
+:$\openint q p \subseteq U$
+Thus by [[Set Intersection Preserves Subsets/Corollary]]:
+:$\openint q p \cap W \subseteq \map f x$
+By [[Axiom:Axiom of Choice|Axiom of Choice]] define a [[Definition:Mapping|mapping]] $f_1: W \to Q$:
+:$\forall x \in W: x \in \map {f_1} x \land \map {f_1} x \cap W \subseteq \map f x$
+By definitions of [[Definition:Image of Subset under Relation|image of set]] and [[Definition:Image of Mapping|image of mapping]]:
+:$\forall A \in \Img {f_1}: \exists x \in W: x \in f_1^{-1} \sqbrk {\set A}$
+By [[Axiom:Axiom of Choice|Axiom of Choice]] define a [[Definition:Mapping|mapping]] $c: \Img {f_1} \to W$:
+:$\forall A \in \Img {f_1}: \map c A \in f_1^{-1} \sqbrk {\set A}$
+Define a [[Definition:Mapping|mapping]] $g: \Img {f_1} \to \CC$:
+:$g := f \circ c$
+Define $\GG = \Img g$.
+Thus $\GG \subseteq \CC$
+By definition of [[Definition:Image of Mapping|image of mapping]]:
+:$\Img {f_1} \subseteq Q$
+By [[Subset of Countable Set is Countable]];
+:$\Img {f_1}$ is [[Definition:Countable Set|countable]].
+By [[Surjection iff Cardinal Inequality]]:
+:$\size {\Img g} \le \size {\Img {f_1} }$
+Thus by [[Countable iff Cardinality not greater than Aleph Zero]]:
+:$\GG$ is [[Definition:Countable Set|countable]].
+It remains to prove that
+:$\GG$ is a [[Definition:Cover of Set|cover]] for $W$.
+Let $x \in W$.
+By definition of $f_1$:
+:$ x \in \map {f_1} x \land \map {f_1} x \cap W \subseteq \map f x$
+By definition of [[Definition:Image of Mapping|image of mapping]]:
+:$\map {f_1} x \in \Img {f_1}$
+Then by definition of $c$:
+:$y := \map c {\map {f_1} x} \in f_1^{-1} \sqbrk {\set {\map {f_1} x} }$
+By definition of $f_1$:
+:$y \in \map {f_1} y \land \map {f_1} y \cap W \subseteq \map f y$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$\map {f_1} y = \map {f_1} x$
+Then by definitions of [[Definition:Subset|subset]] and [[Definition:Set Intersection|intersection]]:
+:$x \in \map f y$
+By definition of [[Definition:Composition of Mappings|composition of mappings]]:
+:$\map f y = \map g {\map {f_1} x}$
+By definition of [[Definition:Image of Mapping|image of mapping]]:
+:$\map g {\map {f_1} x} \in \GG$
+Thus by definition of [[Definition:Union of Set of Sets|union]]:
+:$x \in \bigcup \GG$
+Thus the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0}
+Tags: Examples of Bessel's Equation
+
+\begin{theorem}
+The special case of [[Definition:Bessel's Equation|Bessel's equation]]:
+:$(1): \quad x^2 y'' + x y' + \left({x^2 - \dfrac 1 4}\right) y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 \dfrac {\sin x} {\sqrt x} + C_2 \dfrac {\cos x} {\sqrt x}$
+\end{theorem}
+
+\begin{proof}
+=== [[Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0/Particular Solution|Particular Solution]] ===
+{{:Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0/Particular Solution}}{{qed|lemma}}
+$(1)$ can be expressed as:
+:$(2): \quad y'' + \dfrac 1 x y' + \paren {1 - \dfrac 1 {4 x^2} } y = 0$
+which is in the form:
+:$y'' + \map P x y' + \map Q x y = 0$
+where:
+:$\map P x = \dfrac 1 x$
+:$\map Q x = 1 - \dfrac 1 {4 x^2}$
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int \dfrac 1 x \rd x
+      | c = 
+}}
+{{eqn | r = \ln x
+      | c = [[Primitive of Reciprocal]]
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{-\ln x}
+      | c = 
+}}
+{{eqn | r = \frac 1 x
+      | c =  
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \dfrac x {\sin^2 x} \frac 1 x \rd x
+      | c = 
+}}
+{{eqn | r = \int \csc^2 x \rd x
+      | c = 
+}}
+{{eqn | r = \cot x 
+      | c = [[Primitive of Square of Cosecant Function|Primitive of $\csc^2 x$]]
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = \paren {-\cot x} \dfrac {\sin x} {\sqrt x}
+      | c = 
+}}
+{{eqn | r = -\dfrac {\cos x} {\sqrt x}
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$y = C_1 \dfrac {\sin x} {\sqrt x} + C_2 \dfrac {\cos x} {\sqrt x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Bessel's Equation/x^2 y'' + x y' + (x^2 - (1 over 4)) y = 0/Particular Solution}
+Tags: Examples of Bessel's Equation
+
+\begin{theorem}
+The special case of [[Definition:Bessel's Equation|Bessel's equation]]:
+:$(1): \quad x^2 y'' + x y' + \left({x^2 - \dfrac 1 4}\right) y = 0$
+has a [[Definition:Particular Solution|particular solution]]:
+:$y = \dfrac {\sin x} {\sqrt x}$
+\end{theorem}
+
+\begin{proof}
+Note that:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = \frac {\sin x} {\sqrt x}
+      | c = 
+}}
+{{eqn | r = x^{-1/2} \sin x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}'
+      | r = x^{-1/2} \cos x - \frac 1 2 x^{-3/2} \sin x 
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}''
+      | r = -x^{-1/2} \sin x - \frac 1 2 x^{-3/2} \cos x - \frac 1 2 x^{-3/2} \cos x + \frac 3 4 x^{-5/2} \sin x 
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | r = \left({\frac 3 4 x^{-5/2} - x^{-1/2} }\right) \sin x - x^{-3/2} \cos x 
+      | c = 
+}}
+{{eqn | r = \left({\frac 3 {4 x^2} - 1}\right) x^{-1/2} \sin x - x^{-3/2} \cos x 
+      | c = 
+}}
+{{end-eqn}}
+Inserting into $(1)$, and gradually simplifying:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = x^2 \left({\left({\frac 3 {4 x^2} - 1}\right) x^{-1/2} \sin x - x^{-3/2} \cos x}\right) + x \left({x^{-1/2} \cos x - \frac 1 2 x^{-3/2} \sin x}\right) + \left({x^2 - \dfrac 1 4}\right) x^{-1/2} \sin x
+      | c = 
+}}
+{{eqn | r = \left({\frac 3 4 - x^2}\right) x^{-1/2} \sin x - x^{1/2} \cos x + x^{1/2} \cos x - \frac 1 2 x^{-1/2} \sin x + x^{3/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x
+      | c = 
+}}
+{{eqn | r = \frac 3 4 x^{-1/2} \sin x - x^{3/2} \sin x - \frac 1 2 x^{-1/2} \sin x + x^{3/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x
+      | c = 
+}}
+{{eqn | r = \frac 3 4 x^{-1/2} \sin x - \frac 1 2 x^{-1/2} \sin x - \dfrac 1 4 x^{-1/2} \sin x
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+hence demonstrating that:
+:$y_1 = \dfrac {\sin x} {\sqrt x}$
+is a [[Definition:Particular Solution|particular solution]] of $(1)$.
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/(x - 1) y'' - y' + y = 0}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad \paren {x - 1} y'' - x y' + y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 x + C_2 e^x$
+\end{theorem}
+
+\begin{proof}
+Note that:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}'
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}''
+      | r = 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+and so by inspection:
+:$y_1 = x$
+is a [[Definition:Particular Solution|particular solution]] of $(1)$.
+$(1)$ can be expressed as:
+:$(2): \quad y'' - \dfrac x {x - 1} y' + \dfrac 1 {x - 1} y = 0$
+which is in the form:
+:$y'' + \map P x y' + \map Q x y = 0$
+where:
+:$\map P x = -\dfrac x {x - 1}$
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int \paren {-\dfrac x {x - 1} } \rd x
+      | c = 
+}}
+{{eqn | r = -x - \map \ln {x - 1}
+      | c = [[Primitive of x over a x + b|Primitive of $\dfrac x {a x + b}$]]
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{x + \map \ln {x - 1} }
+      | c = 
+}}
+{{eqn | r = e^x \paren {x - 1}
+      | c =  
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \dfrac 1 {x^2} e^x \paren {x - 1} \rd x
+      | c = 
+}}
+{{eqn | r = \int \paren {\dfrac {e^x} x - \dfrac {e^x} {x^2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \dfrac {e^x} x \rd x - \int \dfrac {e^x} {x^2} \rd x
+      | c = 
+}}
+{{eqn | r = \int \dfrac {e^x} x \rd x + \frac {e^x} x - \int \frac {e^x} x \rd x
+      | c = [[Primitive of Exponential of a x over Power of x|Primitive of $\dfrac {e^{a x} } {x^n}$]]
+}}
+{{eqn | r = \frac {e^x} x
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = \paren {\frac {e^x} x} x
+      | c = 
+}}
+{{eqn | r = e^x
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$y = C_1 x + C_2 e^x$
+{{qed}}
+</onlyinclude>
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/x^2 y'' + 2 x y' - 2 y = 0}
+Tags: Examples of Cauchy-Euler Equation, Linear Second Order ODE: x^2 y'' + 2 x y' - 2 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x^2 y'' + 2 x y' - 2 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 x + \dfrac {C_2} {x^2}$
+\end{theorem}<|endoftext|>
+\section{Second Order ODE/y'' - x f(x) y' + f(x) y = 0}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - x \, \map f x y' + \map f x y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\displaystyle y = C_1 x + C_2 x \int x^{-2} e^{\int x \, \map f x \rd x} \rd x$
+\end{theorem}
+
+\begin{proof}
+Note that:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}'
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}''
+      | r = 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = y'' - x \, \map f x y' + \map f x y
+      | r = 0 - x \, \map f x 1 + \map f x x
+      | c = 
+}}
+{{eqn | r = 0
+      | c = 
+}}
+{{end-eqn}}
+and so it has been demonstrated that:
+:$y_1 = x$
+is a [[Definition:Particular Solution|particular solution]] of $(1)$.
+$(1)$ is in the form:
+:$y'' + \map P x y' + \map Q x y = 0$
+where:
+:$\map P x = -x \, \map f x$
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int \paren {-x \, \map f x} \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{-\int \paren {-x \, \map f x} \rd x}
+      | c = 
+}}
+{{eqn | r = e^{\int x \, \map f x \rd x}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \dfrac 1 {x^2} e^{\int x \, \map f x \rd x} \rd x
+      | c = as $y_1 = x$
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = x \int x^{-2} e^{\int x \, \map f x \rd x} \rd x
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$\displaystyle y = C_1 x + C_2 x \int x^{-2} e^{\int x \, \map f x \rd x} \rd x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/x y'' - (2 x + 1) y' + (x + 1) y = 0}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x y'' - \paren {2 x + 1} y' + \paren {x + 1} y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 e^x + C_2 x^2 e^x$
+\end{theorem}
+
+\begin{proof}
+Note that:
+:$x - \paren {2 x + 1} + \paren {x + 1} = 0$
+so if $y'' = y' = y$ we find that $(1)$ is satisfied.
+So:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = e^x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}'
+      | r = e^x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}''
+      | r = e^x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{end-eqn}}
+and so:
+:$y_1 = e^x$
+is a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(1)$.
+$(1)$ can be expressed as:
+:$(2): \quad y'' - \dfrac {2 x + 1} x y' + \dfrac {x + 1} x y = 0$
+which is in the form:
+:$y'' + \map P x y' + \map Q x y = 0$
+where:
+:$\map P x = -\dfrac {2 x + 1} x$
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int \paren {-\dfrac {2 x + 1} x} \rd x
+      | c = 
+}}
+{{eqn | r = \int \paren {-2 - \dfrac 1 x} \rd x
+      | c = 
+}}
+{{eqn | r = -2 x - \ln x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{-\paren {-2 x - \ln x} }
+      | c = 
+}}
+{{eqn | r = e^{2 x + \ln x}
+      | c = 
+}}
+{{eqn | r = e^{2 x} e^{\ln x}
+      | c = 
+}}
+{{eqn | r = x e^{2 x}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \dfrac 1 {e^{2 x} } x e^{2 x} \rd x
+      | c = as $y_1 = e^x$
+}}
+{{eqn | r = \int x \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^2} 2
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = \frac {x^2} 2 e^x
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$y = C_1 e^x + k \dfrac {x^2} 2 e^x$
+and so setting $C_2 = \dfrac k 2$:
+:$y = C_1 e^x + C_2 x^2 e^x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/y'' - f(x) y' + (f(x) - 1) y = 0}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - \map f x y' + \paren {\map f x - 1} y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$\displaystyle y = C_1 e^x + C_2 e^x \int e^{-2 x + \int \map f x \rd x} \rd x$
+\end{theorem}
+
+\begin{proof}
+Note that:
+:$1 - \map f x + \paren {\map f x - 1} = 0$
+so if $y'' = y' = y$ we find that $(1)$ is satisfied.
+So:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = e^x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}'
+      | r = e^x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}''
+      | r = e^x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{end-eqn}}
+and so:
+:$y_1 = e^x$
+is a [[Definition:Particular Solution|particular solution]] of $(1)$.
+$(1)$ is in the form:
+:$y'' + \map P x y' + \map Q x y = 0$
+where:
+:$\map P x = -\map f x$
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int \paren {-\map f x} \rd x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{-\int \paren {-\map f x} \rd x}
+      | c = 
+}}
+{{eqn | r = e^{\int \map f x \rd x}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \dfrac 1 {e^{2 x} } e^{\int \map f x \rd x} \rd x
+      | c = as $y_1 = e^x$
+}}
+{{eqn | r = \int e^{-2 x + \int \map f x \rd x} \rd x
+      | c = as $y_1 = e^x$
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = e^x \int e^{-2 x + \int \map f x \rd x} \rd x
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$\displaystyle y = C_1 e^x + C_2 e^x \int e^{-2 x + \int \map f x \rd x} \rd x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation}
+Tags: Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+Let:
+:$(1): \quad y'' + p y' + q y = 0$
+be a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Then:
+:$y = e^{m_1 x}$ is a [[Definition:Solution to Differential Equation|solution]] to $(1)$
+{{iff}}
+:$m_1$ is a [[Definition:Root of Polynomial|root]] of the [[Definition:Auxiliary Equation|auxiliary equation]] $m^2 + p m + q = 0$
+\end{theorem}
+
+\begin{proof}
+Consider the equation:
+: $y = e^{m_1 x}$
+[[Definition:Differentiation|Differentiating]] {{WRT|Differentiation}} $x$:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = e^{m_1 x}
+      | c = 
+}}
+{{eqn | l = y'
+      | r = m_1 e^{m_1 x}
+      | c = 
+}}
+{{eqn | l = y''
+      | r = {m_1}^2 e^{m_1 x}
+      | c = 
+}}
+{{end-eqn}}
+=== Sufficient Condition ===
+Let $y = e^{m_1 x}$ be a [[Definition:Solution to Differential Equation|solution]] to $(1)$.
+Substituting for $y$ and its [[Definition:Derivative|derivatives]] in $(1)$:
+{{begin-eqn}}
+{{eqn | l = y'' + p y' + q y
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {m_1}^2 e^{m_1 x} + p m_1 e^{m_1 x} + q e^{m_1 x} 
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren { {m_1}^2 + p m_1 + q} e^{m_1 x}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {m_1}^2 + p m_1 + q
+      | r = 0
+      | c = as $e^{m_1 x}$ is never zero
+}}
+{{end-eqn}}
+That is, by definition, $m_1$ is a [[Definition:Root of Polynomial|root]] of $m^2 + p m + q = 0$.
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $m_1$ be a [[Definition:Root of Polynomial|root]] of $m^2 + p m + q = 0$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = {m_1}^2 + p m_1 + q
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren { {m_1}^2 + p m_1 + q} e^{m_1 x}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {m_1}^2 e^{m_1 x} + p m_1 e^{m_1 x} + q e^{m_1 x} 
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y'' + p y' + q y
+      | r = 0
+      | c = substituting for $y$ and its [[Definition:Derivative|derivatives]]
+}}
+{{end-eqn}}
+Thus $y = e^{m_1 x}$ satisfies $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Solution of Constant Coefficient Homogeneous LSOODE}
+Tags: Constant Coefficient Homogeneous LSOODEs, Solution of Constant Coefficient Homogeneous LSOODE
+
+\begin{theorem}
+Let:
+:$(1): \quad y'' + p y' + q y = 0$
+be a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Let $m_1$ and $m_2$ be the [[Definition:Root of Polynomial|roots]] of the [[Definition:Auxiliary Equation|auxiliary equation]] $m^2 + p m + q = 0$.
+\end{theorem}
+
+\begin{proof}
+=== [[Solution of Constant Coefficient Homogeneous LSOODE/Real Roots of Auxiliary Equation|Real Roots of Auxiliary Equation]] ===
+{{:Solution of Constant Coefficient Homogeneous LSOODE/Real Roots of Auxiliary Equation}}{{qed|lemma}}
+=== [[Solution of Constant Coefficient Homogeneous LSOODE/Equal Real Roots of Auxiliary Equation|Equal Real Roots of Auxiliary Equation]] ===
+{{:Solution of Constant Coefficient Homogeneous LSOODE/Equal Real Roots of Auxiliary Equation}}{{qed|lemma}}
+=== [[Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation|Complex Roots of Auxiliary Equation]] ===
+{{:Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation}}{{qed}}
+\end{proof}<|endoftext|>
+\section{Solution of Constant Coefficient Homogeneous LSOODE/Real Roots of Auxiliary Equation}
+Tags: Solution of Constant Coefficient Homogeneous LSOODE
+
+\begin{theorem}
+{{:Solution of Constant Coefficient Homogeneous LSOODE}}
+Let $p^2 > 4 q$.
+Then $(1)$ has the [[Definition:General Solution|general solution]]:
+:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Auxiliary Equation|auxiliary equation]] of $(1)$:
+:$(2): \quad m^2 + p m + q$
+Let $p^2 > 4 q$.
+From [[Solution to Quadratic Equation with Real Coefficients]], $(2)$ has two [[Definition:Real Number|real]] [[Definition:Root of Polynomial|roots]]:
+{{begin-eqn}}
+{{eqn | l = m_1
+      | r = -\frac p 2 + \sqrt {\frac {p^2} 4 - q}
+}}
+{{eqn | l = m_2
+      | r = -\frac p 2 - \sqrt {\frac {p^2} 4 - q}
+}}
+{{end-eqn}}
+As $p^2 > 4 q$ we have that:
+:$\sqrt {\dfrac {p^2} 4 - q} \ne 0$
+and so:
+:$m_1 \ne m_2$
+From [[Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation]]:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = e^{m_1 x}
+}}
+{{eqn | l = y_2
+      | r = e^{m_2 x}
+}}
+{{end-eqn}}
+are both [[Definition:Particular Solution|particular solutions]] to $(1)$.
+We also have that:
+{{begin-eqn}}
+{{eqn | l = \frac {y_1} {y_2}
+      | r = \frac {e^{m_1 x} } {e^{m_2 x} }
+}}
+{{eqn | r = e^{\paren {m_1 - m_2} x}
+}}
+{{eqn | o = \ne
+      | r = 0
+      | c = as $m_1 \ne m_2$
+}}
+{{end-eqn}}
+Thus $y_1$ and $y_2$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+It follows from [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]] that:
+:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
+is the [[Definition:General Solution|general solution]] to $(1)$.
+\end{proof}<|endoftext|>
+\section{Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation}
+Tags: Solution of Constant Coefficient Homogeneous LSOODE
+
+\begin{theorem}
+{{:Solution of Constant Coefficient Homogeneous LSOODE}}
+Let $p^2 < 4 q$.
+Then $(1)$ has the [[Definition:General Solution|general solution]]:
+:$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$
+where:
+:$m_1 = a + i b$
+:$m_2 = a - i b$
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Auxiliary Equation|auxiliary equation]] of $(1)$:
+:$(2): \quad m^2 + p m + q$
+Let $p^2 < 4 q$.
+From [[Solution to Quadratic Equation with Real Coefficients]], $(2)$ has two [[Definition:Complex Number|complex]] [[Definition:Root of Polynomial|roots]]:
+{{begin-eqn}}
+{{eqn | l = m_1
+      | r = -\frac p 2 + i \sqrt {q - \frac {p^2} 4}
+}}
+{{eqn | l = m_2
+      | r = -\frac p 2 - i \sqrt {q - \frac {p^2} 4}
+}}
+{{end-eqn}}
+As $p^2 < 4 q$ we have that:
+:$\sqrt {q - \dfrac {p^2} 4} \ne 0$
+and so:
+:$m_1 \ne m_2$
+Let:
+{{begin-eqn}}
+{{eqn | l = m_1
+      | r = a + i b
+}}
+{{eqn | l = m_2
+      | r = a - i b
+}}
+{{end-eqn}}
+where $a = -\dfrac p 2$ and $b = \sqrt {q - \dfrac {p^2} 4}$.
+From [[Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation]]:
+{{begin-eqn}}
+{{eqn | l = y_a
+      | r = e^{m_1 x}
+}}
+{{eqn | l = y_b
+      | r = e^{m_2 x}
+}}
+{{end-eqn}}
+are both [[Definition:Particular Solution|particular solutions]] to $(1)$.
+We can manipulate $y_a$ and $y_b$ into the following forms:
+{{begin-eqn}}
+{{eqn | l = y_a
+      | r = e^{m_1 x}
+      | c = 
+}}
+{{eqn | r = e^{\paren {a + i b} x}
+      | c = 
+}}
+{{eqn | r = e^{a x} e^{i b x}
+      | c = 
+}}
+{{eqn | n = 3
+      | r = e^{a x} \paren {\cos b x + i \sin b x}
+      | c = [[Euler's Formula]]
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = y_b
+      | r = e^{m_2 x}
+      | c = 
+}}
+{{eqn | r = e^{\paren {a - i b} x}
+      | c = 
+}}
+{{eqn | r = e^{a x} e^{-i b x}
+      | c = 
+}}
+{{eqn | n = 4
+      | r = e^{a x} \paren {\cos b x - i \sin b x}
+      | c = [[Euler's Formula/Corollary|Euler's Formula: Corollary]]
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_a + y_b
+      | r = e^{a x} \paren {\cos b x + i \sin b x} + e^{a x} \paren {\cos b x - i \sin b x}
+      | c = adding $(3)$ and $(4)$
+}}
+{{eqn | r = 2 e^{a x} \cos b x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y_a + y_b} 2
+      | r = e^{a x} \cos b x
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = y_b - y_a
+      | r = e^{a x} \paren {\cos b x - i \sin b x} - e^{a x} \paren {\cos b x + i \sin b x}
+      | c = subtracting $(4)$ from $(3)$
+}}
+{{eqn | r = 2 e^{a x} \sin b x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {y_b - y_a} 2
+      | r = e^{a x} \sin b x
+      | c = 
+}}
+{{end-eqn}}
+Let:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = \frac {y_a + y_b} 2
+}}
+{{eqn | r = e^{a x} \cos b x
+      | c = 
+}}
+{{eqn | l = y_2
+      | r = \frac {y_b - y_a} 2
+}}
+{{eqn | r = e^{a x} \sin b x
+      | c = 
+}}
+{{end-eqn}}
+We have that:
+{{begin-eqn}}
+{{eqn | l = \frac {y_1} {y_2}
+      | r = \frac {e^{a x} \cos b x} {e^{a x} \sin b x}
+}}
+{{eqn | r = \cot b x
+}}
+{{end-eqn}}
+As $\cot b x$ is not zero for all $x$, $y_1$ and $y_2$ are [[Definition:Linearly Independent Real Functions|linearly independent]].
+From [[Linear Combination of Solutions to Homogeneous Linear 2nd Order ODE]]:
+:$y_1 = \dfrac {y_a + y_b} 2$
+:$y_2 = \dfrac {y_b - y_b} 2$
+are both [[Definition:Particular Solution|particular solutions]] to $(1)$.
+It follows from [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]] that:
+:$y = C_1 e^{a x} \cos b x + C_2 e^{a x} \sin b x$
+or:
+:$y = e^{a x} \paren {C_1 \cos b x + C_2 \sin b x}$
+is the [[Definition:General Solution|general solution]] to $(1)$.
+\end{proof}<|endoftext|>
+\section{Solution of Constant Coefficient Homogeneous LSOODE/Equal Real Roots of Auxiliary Equation}
+Tags: Solution of Constant Coefficient Homogeneous LSOODE
+
+\begin{theorem}
+{{:Solution of Constant Coefficient Homogeneous LSOODE}}
+Let $p^2 = 4 q$.
+Then $(1)$ has the [[Definition:General Solution|general solution]]:
+:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Auxiliary Equation|auxiliary equation]] of $(1)$:
+:$(2): \quad m^2 + p m + q$
+Let $p^2 = 4 q$.
+From [[Solution to Quadratic Equation with Real Coefficients]], $(2)$ has one (repeated) [[Definition:Root of Polynomial|root]], that is:
+:$m_1 = m_2 = -\dfrac p 2$
+From [[Exponential Function is Solution of Constant Coefficient Homogeneous LSOODE iff Index is Root of Auxiliary Equation]]:
+:$y_1 = e^{m_1 x}$
+is a [[Definition:Particular Solution|particular solution]] to $(1)$.
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int p \rd x
+      | c = 
+}}
+{{eqn | r = p x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{-p x}
+      | c = 
+}}
+{{eqn | r = e^{2 m_1 x}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \dfrac 1 {e^{2 m_1 x} } e^{2 m_1 x} \rd x
+      | c = as $y_1 = e^{m_1 x}$
+}}
+{{eqn | r = \int \rd x
+      | c = 
+}}
+{{eqn | r = x
+      | c = 
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = x e^{m_1 x}
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + y' - 6 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + y' - 6 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{2 x} + C_2 e^{-3 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + m - 6 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = -3$
+:$m_2 = 2$
+These are [[Definition:Real Number|real]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{2 x} + C_2 e^{-3 x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 2 y' + y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 2 y' + y = 0$
+has the [[Definition:General Solution|general solution]]:
+:$y = C_1 e^{-x} + C_2 x e^{-x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad m^2 + 2 m + 1 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = m_2 = -1$
+These are [[Definition:Real Number|real]] and equal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution|general solution]] of $(1)$ is:
+:$y = C_1 e^{-x} + C_2 x e^{-x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 8 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 8 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \cos 2 \sqrt 2 x + C_2 \sin 2 \sqrt 2 x$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + 8 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 2 \sqrt 2 i$
+:$m_2 = -2 \sqrt 2 i$
+These are [[Definition:Complex Number|complex]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 \cos 2 \sqrt 2 x + C_2 \sin 2 \sqrt 2 x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/2 y'' - 4 y' + 8 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad 2 y'' - 4 y + 8 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^x \paren {C_1 \cos \sqrt 3 x + C_2 \sin \sqrt 3 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Let $(1)$ be written in the form:
+:$y'' - 2 y + 4 y = 0$
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 - 2 m + 4 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 1 + \sqrt 3 i$
+:$m_2 = 1 - \sqrt 3 i$
+These are [[Definition:Complex Number|complex]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution|general solution]] of $(1)$ is:
+:$y = e^x \paren {C_1 \cos \sqrt 3 x + C_2 \sin \sqrt 3 x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 9 y' + 20 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 9 y' + 20 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{4 x} + C_2 e^{5 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 - 9 m + 20 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 4$
+:$m_2 = 5$
+These are [[Definition:Real Number|real]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{4 x} + C_2 e^{5 x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/2 y'' + 2 y' + 3 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad 2 y'' + 2 y' + 3 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^{-x/2} \paren {C_1 \cos \dfrac {\sqrt 5} 2 x + C_2 \sin \dfrac {\sqrt 5} 2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: 2 m^2 + 2 m + 3 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = -\dfrac 1 2 + \dfrac {\sqrt 5} 2 i$
+:$m_2 = -\dfrac 1 2 - \dfrac {\sqrt 5} 2 i$
+These are [[Definition:Complex Number|complex]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution|general solution]] of $(1)$ is:
+:$y = e^{-x/2} \paren {C_1 \cos \dfrac {\sqrt 5} 2 x + C_2 \sin \dfrac {\sqrt 5} 2 x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Condition for Solutions to Constant Coefficient Homogeneous LSOODE to tend to Zero}
+Tags: Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+Let:
+:$(1): \quad y'' + p y' + q y = 0$
+be a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Let the [[Definition:General Solution|general solution]] to $(1)$ be $\map y {x, C_1, C_2}$.
+Then:
+:$\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
+{{iff}}
+:$p$ and $q$ are both [[Definition:Strictly Positive Real Number|strictly positive]].
+\end{theorem}
+
+\begin{proof}
+By [[Solution of Constant Coefficient Homogeneous LSOODE]], $y$ is in one of the following forms:
+:$y = \begin{cases}
+C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\
+ & \\
+C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\
+ & \\
+e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q
+\end{cases}$
+where:
+:$m_1$ and $m_2$ are the [[Definition:Root of Polynomial|roots]] of the [[Definition:Auxiliary Equation|auxiliary equation]] $m^2 + p m + q = 0$
+:$a + i b = m_1$
+:$a - i b = m_2$
+=== Sufficient Condition ===
+Let:
+:$\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
+Let $y$ be of the form:
+:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
+Then it follows that $m_1 < 0$ and $m_2 < 0$.
+From [[Sum of Roots of Quadratic Equation]]:
+:$p = -\paren {m_1 + m_2}$
+from which it follows that:
+:$p > 0$
+From [[Product of Roots of Quadratic Equation]]:
+:$q = m_1 m_2$
+from which it follows that:
+:$q > 0$
+{{qed|lemma}}
+Let $y$ be of the form:
+:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
+From [[Limit at Infinity of Polynomial over Complex Exponential]]:
+:$\displaystyle \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$
+{{iff}} $m_1 < 0$.
+Thus it follows that $m_1 < 0$.
+Again from [[Sum of Roots of Quadratic Equation]]:
+:$p = -\paren {2 m_1}$
+from which it follows that:
+:$p > 0$
+From [[Product of Roots of Quadratic Equation]]:
+:$q = m_1^2$
+from which it follows that:
+:$q > 0$
+{{qed|lemma}}
+Let $y$ be of the form:
+:$y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$
+Thus it follows that $a < 0$.
+From [[Sum of Roots of Quadratic Equation]]:
+{{begin-eqn}}
+{{eqn | l = p
+      | r = -\paren {a + i b + a - i b}
+      | c = [[Sum of Roots of Quadratic Equation]]
+}}
+{{eqn | r = -2 a
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = p
+      | o = >
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = q
+      | r = \paren {a + i b} \paren {a - i b}
+      | c = [[Product of Roots of Quadratic Equation]]
+}}
+{{eqn | r = a^2 + b^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = q
+      | o = >
+      | r = 0
+      | c = whatever $a$ and $b$ are
+}}
+{{end-eqn}}
+{{qed|lemma}}
+Thus it is seen that in all three cases:
+:$\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
+implies that $p$ and $q$ are both [[Definition:Strictly Positive Real Number|strictly positive]].
+=== Necessary Condition ===
+Let $p$ and $q$ both be [[Definition:Strictly Positive Real Number|strictly positive]].
+Let $p^2 > 4 q$.
+Then both $m_1$ and $m_2$ are [[Definition:Real Number|real]] and unequal.
+Thus from [[Solution of Constant Coefficient Homogeneous LSOODE]]:
+:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
+From [[Product of Roots of Quadratic Equation]]:
+:$m_1 m_2 = q > 0$
+and so either $m_1$ and $m_2$ are either both [[Definition:Strictly Positive Real Number|strictly positive]] or [[Definition:Strictly Negative Real Number|strictly negative]].
+It follows that $m_1 + m_2$ is also either [[Definition:Strictly Positive Real Number|strictly positive]] or [[Definition:Strictly Negative Real Number|strictly negative]].
+From [[Sum of Roots of Quadratic Equation]]:
+:$-\paren {m_1 + m_2} = p > 0$
+and so $m_1 + m_2 < 0$
+Hence both $m_1$ and $m_2$ are [[Definition:Strictly Negative Real Number|strictly negative]].
+It follows that: 
+:$\displaystyle \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 e^{m_2 x} = 0$
+{{qed|lemma}}
+Let $p^2 = 4 q$.
+Then from [[Solution to Quadratic Equation with Real Coefficients]]:
+:$m_1 = m_2 = -\frac p 2$
+and so:
+:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
+From [[Limit at Infinity of Polynomial over Complex Exponential]]:
+:$\displaystyle \lim_{x \mathop \to \infty} C_2 x e^{m_1 x} = 0$
+{{iff}} $m_1 < 0$.
+It follows that: 
+:$\displaystyle \lim_{x \mathop \to \infty} C_1 e^{m_1 x} + C_2 x e^{m_1 x} = 0$
+{{qed|lemma}}
+Let $p^2 < 4 q$.
+Then both $m_1$ and $m_2$ are [[Definition:Complex Number|complex]] and unequal:
+:$m_1 = a + i b = -\dfrac p 2 + i \dfrac {\sqrt {4 q - p^2} } 2$
+:$m_2 = a + i b = -\dfrac p 2 - i \dfrac {\sqrt {4 q - p^2} } 2$
+Then from [[Solution of Constant Coefficient Homogeneous LSOODE]]:
+:$y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$
+As $p > 0$ it follows that $a < 0$.
+Thus it follows that: 
+:$\displaystyle \lim_{x \mathop \to \infty} e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} = 0$
+{{qed|lemma}}
+Thus in all cases:
+:$\displaystyle \lim_{x \mathop \to \infty} \map y {x, C_1, C_2} = 0$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Solution to Constant Coefficient Homogeneous LSOODE is also Solution}
+Tags: Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+Let:
+:$(1): \quad y'' + p y' + q y = 0$
+be a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Let $\map y x$ be a [[Definition:Particular Solution|particular solution]] of $(1)$.
+Then its [[Definition:Derivative|derivative]] $\map {y'} x$ is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+\end{theorem}
+
+\begin{proof}
+By [[Solution of Constant Coefficient Homogeneous LSOODE]], $y$ is in one of the following forms:
+:$y = \begin{cases}
+C_1 e^{m_1 x} + C_2 e^{m_2 x} & : p^2 > 4 q \\
+ & \\
+C_1 e^{m_1 x} + C_2 x e^{m_2 x} & : p^2 = 4 q \\
+ & \\
+e^{a x} \paren {C_1 \sin b x + C_2 \cos b x} & : p^2 < 4 q
+\end{cases}$
+Let:
+:$y = A e^{m_1 x} + B e^{m_2 x}$
+where $A, B \in \R$.
+Then:
+:$y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$
+which is also in the form:
+:$y = C_1 e^{m_1 x} + C_2 e^{m_2 x}$
+and so is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+{{qed|lemma}}
+Let:
+:$y = A e^{m_1 x} + B x e^{m_1 x}$
+where $A, B \in \R$.
+Then:
+{{begin-eqn}}
+{{eqn | l = y'
+      | r = m_1 A e^{m_1 x} + B e^{m_1 x} + B m_1 x e^{m_1 x}
+      | c = 
+}}
+{{eqn | r = \paren {m_1 A + B} e^{m_1 x} + B m_1 x e^{m_1 x}
+      | c = 
+}}
+{{end-eqn}}
+which is also in the form:
+:$y = C_1 e^{m_1 x} + C_2 x e^{m_1 x}$
+and so is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+{{qed|lemma}}
+Let:
+:$y = e^{a x} \left({A \sin b x + B \cos b x}\right)$
+where $A, B \in \R$.
+Then:
+:$y' = m_1 A e^{m_1 x} + m_2 B e^{m_2 x}$
+{{begin-eqn}}
+{{eqn | l = y'
+      | r = a e^{a x} \paren {A \sin b x + B \cos b x} + e^{a x} \paren {A b \cos b x - B b \sin b x}
+      | c = 
+}}
+{{eqn | r = e^{a x} \paren {a A \sin b x + a B \cos b x + A b \cos b x - B b \sin b x}
+      | c = 
+}}
+{{eqn | r = e^{a x} \paren {\paren {a B + A b} \cos b x + \paren {a A - b B} \sin b x}
+      | c = 
+}}
+{{end-eqn}}
+which is also in the form:
+:$y = e^{a x} \paren {C_1 \sin b x + C_2 \cos b x}$
+and so is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+{{qed|lemma}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE}
+Tags: Cauchy-Euler Equation
+
+\begin{theorem}
+Consider the [[Definition:Cauchy-Euler Equation|Cauchy-Euler equation]]:
+:$(1): \quad x^2 \dfrac {\d^2 y} {\d x^2} + p x \dfrac {\d y} {\d x} + q y = 0$
+By making the substitution:
+:$x = e^t$
+it is possible to convert $(1)$ into a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t} + q y = 0$
+\end{theorem}
+
+\begin{proof}
+We have:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = e^t
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | ll=\leadsto
+      | l = \frac {\d x} {\d t}
+      | r = e^t
+      | c = 
+}}
+{{eqn | n = 1
+      | r = x
+      | c = 
+}}
+{{eqn | ll=\leadsto
+      | l = \frac {\d t} {\d x}
+      | r = \frac 1 x
+      | c = [[Derivative of Inverse Function]]
+}}
+{{eqn | n = 2
+      | r = e^{-t}
+      | c = 
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \frac {\d y} {\d x}
+      | r = \frac {\d y} {\d t} \frac {\d t} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \frac {\d y} {\d t} e^{-t}
+      | c = from $(2)$
+}}
+{{end-eqn}}
+and:
+{{begin-eqn}}
+{{eqn | l = \frac {\d^2 y} {\d x^2}
+      | r = \frac {\d} {\d x} \paren {\frac {\d y} {\d x} }
+      | c = 
+}}
+{{eqn | r = \frac {\d} {\d t} \paren {\frac {\d y} {\d x} } \frac {\d t} {\d x}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = \frac {\d} {\d t} \paren {\frac {\d y} {\d t} e^{-t} } e^{-t}
+      | c = from $(2)$
+}}
+{{eqn | r = e^{-t} \paren {e^{-t} \frac {\d^2 y} {\d t^2} - e^{-t} \frac {\d y} {\d t} }
+      | c = [[Product Rule]]
+}}
+{{eqn | r = e^{- 2 t} \paren {\frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} }
+      | c = 
+}}
+{{end-eqn}}
+Substituting back into $(1)$:
+{{begin-eqn}}
+{{eqn | l = x^2 \dfrac {\d^2 y} {\d x^2} + p x \dfrac {\d y} {\d x} + q y
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = e^{2 t} \paren {e^{- 2 t} \paren {\frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} } } + p e^t \paren {\frac {\d y} {\d t} e^{-t} } + q y
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \frac {\d^2 y} {\d t^2} - \frac {\d y} {\d t} + p \frac {\d y} {\d t} + q y
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \frac {\d^2 y} {\d t^2} + \paren {p - 1} \frac {\d y} {\d t} + q y
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/x^2 y'' + 3 x y' + 10 y = 0}
+Tags: Examples of Cauchy-Euler Equation
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x^2 y'' + 3 x y' + 10 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = \dfrac 1 x \paren {C_1 \, \map \cos {\ln x^3} + C_2 \, \map \sin {\ln x^3} }$
+\end{theorem}
+
+\begin{proof}
+$(1)$ is an instance of the [[Definition:Cauchy-Euler Equation|Cauchy-Euler Equation]]:
+:$x^2 y'' + p x y' + q y = 0$
+where:
+:$p = 3$
+:$q = 10$
+By [[Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE]], this can be expressed as:
+:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$
+by making the substitution:
+:$x = e^t$
+Hence $(1)$ can be expressed as:
+:$(2): \quad \dfrac {\d^2 y} {\d t^2} + 2 \dfrac {\d y} {\d t^2} + 10 y = 0$
+It can be seen that $(2)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+From [[Linear Second Order ODE/y'' + 2 y' + 10 y = 0|Linear Second Order ODE: $y'' + 2 y' + 10 y = 0$]], this has the [[Definition:General Solution|general solution]]:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = e^{-t} \paren {C_1 \cos 3 t + C_2 \sin 3 t}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 x \paren {C_1 \cos 3 t + C_2 \sin 3 t}
+      | c = substituting $x = e^t$
+}}
+{{eqn | r = \dfrac 1 x \paren {C_1 \, \map \cos {3 \ln x} + C_2 \, \map \sin {3 \ln x} }
+      | c = substituting $t = \ln x$
+}}
+{{eqn | r = \dfrac 1 x \paren {C_1 \, \map \cos {\ln x^3} + C_2 \, \map \sin {\ln x^3} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/2 x^2 y'' + 10 x y' + 8 y = 0}
+Tags: Examples of Cauchy-Euler Equation
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad 2 x^2 y'' + 10 x y' + 8 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 x^{-2} + C_2 x^{-2} \ln x$
+\end{theorem}
+
+\begin{proof}
+Let $(1)$ be rewritten as:
+:$x^2 y'' + 5 x y' + 4 y = 0$
+It can be seen to be an instance of the [[Definition:Cauchy-Euler Equation|Cauchy-Euler Equation]]:
+:$x^2 y'' + p x y' + q y = 0$
+where:
+:$p = 5$
+:$q = 4$
+By [[Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE]], this can be expressed as:
+:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$
+by making the substitution:
+:$x = e^t$
+Hence it can be expressed as:
+:$(2): \quad \dfrac {\d^2 y} {\d t^2} + 4 \dfrac {\d y} {\d t^2} + 4 y = 0$
+From [[Linear Second Order ODE/y'' + 4 y' + 4 y = 0|Linear Second Order ODE: $y'' + 4 y' + 4 y = 0$]], this has the [[Definition:General Solution|general solution]]:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = C_1 e^{-2 t} + C_2 t e^{-2 t}
+      | c = 
+}}
+{{eqn | r = C_1 x^{-2} + C_2 t x^{-2}
+      | c = substituting $x = e^t$
+}}
+{{eqn | r = C_1 x^{-2} + C_2 x^{-2} \ln x
+      | c = substituting $t = \ln x$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/x^2 y'' + 2 x y' - 12 y = 0}
+Tags: Examples of Cauchy-Euler Equation
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad x^2 y'' + 2 x y' - 12 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 x^3 + C_2 x^{-4}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is an instance of the [[Definition:Cauchy-Euler Equation|Cauchy-Euler Equation]]:
+:$x^2 y'' + p x y' + q y = 0$
+where:
+:$p = 2$
+:$q = -12$
+By [[Conversion of Cauchy-Euler Equation to Constant Coefficient Linear ODE]], this can be expressed as:
+:$\dfrac {\d^2 y} {\d t^2} + \paren {p - 1} \dfrac {\d y} {\d t^2} + q y = 0$
+by making the substitution:
+:$x = e^t$
+Hence it can be expressed as:
+:$(2): \quad \dfrac {\d^2 y} {\d t^2} + \dfrac {\d y} {\d t^2} - 12 y = 0$
+From [[Linear Second Order ODE/y'' + y' - 12 y = 0|Linear Second Order ODE: $y'' + y' - 12 y = 0$]], this has the [[Definition:General Solution|general solution]]:
+{{begin-eqn}}
+{{eqn | l = y
+      | r = C_1 e^{3 t} + C_2 e^{-4 t}
+      | c = 
+}}
+{{eqn | r = C_1 x^3 + C_2 x^{-4}
+      | c = substituting $x = e^t$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Image of Set not greater than Cardinality of Set}
+Tags: Cardinals, Mapping Theory
+
+\begin{theorem}
+Let $X, Y$ be [[Definition:Set|sets]].
+Let $f:X \to Y$ be a [[Definition:Mapping|mapping]].
+Let $A$ be a [[Definition:Subset|subset]] of $X$.
+Then $\left\vert{f^\to\left({A}\right)}\right\vert \le \left\vert{A}\right\vert$
+where $\left\vert{A}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $A$.
+\end{theorem}
+
+\begin{proof}
+By definitions of [[Definition:Surjection|surjection]] and [[Definition:Restriction of Mapping|restriction of mapping]]:
+:$F \restriction_A: A \to f^\to\left({A}\right)$ is a [[Definition:Surjection|surjection]]
+Thus by [[Surjection iff Cardinal Inequality]]:
+:$\left\vert{f^\to\left({A}\right)}\right\vert \le \left\vert{A}\right\vert$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Projector has Norm 1}
+Tags: 
+
+\begin{theorem}
+An [[Idempotent Operators|idempotent operator]] $P$ is a [[projector]] on the [[Definition:Hilbert Space|Hilbert Space]] $H$ {{iff}} $P$ has [[Definition:Norm on Bounded Linear Transformation|norm]] $1$:
+:$\displaystyle \norm P \equiv \sup_{x \mathop \in H} \frac {\norm P} {\norm x} = 1$
+\end{theorem}
+
+\begin{proof}
+For all $x \in \Rng P$:
+:$\norm {P \dfrac x {\norm x} } = \dfrac {\norm x} {\norm x} = 1$
+so $\norm P \ge 1$.
+It remains to show that this holds with equality {{iff}} $P$ is a projector.
+First, suppose $P$ is a projector.
+Let $\set {p_1, p_2, \ldots}$ be an [[Definition:Orthonormal Subset|orthonormal]] [[Definition:Basis (Linear Algebra)|basis]] for $\Rng P$.
+Let $\set {q_1, q_2, \ldots}$ be an [[Definition:Orthonormal Subset|orthonormal]] [[Definition:Basis (Linear Algebra)|basis]] for $\Rng P_\perp$.
+Then for any $x \in H$, we can choose scalars $\set {\alpha_1, \alpha_2, \ldots}$ and $\set {\beta_1, \beta_2, \ldots}$ so that:
+:$\displaystyle x = \sum_{i \mathop = 1}^\infty \alpha_i p_i + \sum_{i \mathop = 1}^\infty \beta_i q_i$
+Because the basis vectors are orthogonal, [[Pythagoras's Theorem (Hilbert Space)|Pythagoras's theorem]] shows that:
+:$\displaystyle \norm x^2 = \sum_{i \mathop = 1}^\infty \norm {\alpha_i}^2 + \norm {\beta_i}^2$
+Then $\norm {P x}$ can be expanded thus:
+{{begin-eqn}}
+{{eqn | l = \norm {P x}^2
+      | r = \norm {\sum_{i \mathop = 1}^\infty \alpha_i P p_i + \sum_{i \mathop = 1}^\infty \beta_i P q_i}^2
+}}
+{{eqn| r = \norm {\sum_{i \mathop = 1}^\infty \alpha_i p_i + 0}^2
+}}
+{{eqn | r = \sum_{i \mathop = 1}^\infty \size {\alpha_i}^2
+      | c = [[Pythagoras's Theorem (Hilbert Space)|Pythagoras's theorem]] again
+}}
+{{eqn | o = \le
+      | r = \norm x^2
+}}
+{{end-eqn}}
+Hence $\norm P \le 1$.
+Since it was already shown $\norm P \ge 1$, it follows that $\norm P = 1$.
+Now suppose $P$ is not a projector.
+Then there exists $x \in H$ so that $P x - x$ is not orthogonal to $\Rng P$.
+By writing $x = p + q$ with $p \in \Rng P$ and $q \in \Rng P_\perp$, it follows that:
+:$P x - x = P \paren {p + q} - \paren {p + q} = P q - q$
+By rescaling $x$, we can assume $\norm q = 1$.
+Since $q \in \Rng P_\perp$ but
+:$P q - q = P x - x = \notin \Rng P_\perp$
+it follows that $P q \ne 0$.
+Let $\norm {P q} = c \ne 0$.
+It will be shown that
+:$y = c q + \dfrac 1 c P q$
+satisfies
+:$\dfrac {\norm {P y} } {\norm y} > 1$.
+Notice first that
+{{begin-eqn}}
+{{eqn | l = \norm y^2
+      | r = \norm {c q + \dfrac 1 c P q}^2
+}}
+{{eqn | r = \norm {c q}^2 + \norm {\dfrac 1 c P q}^2
+      | c = [[Pythagoras's Theorem (Hilbert Space)|Pythagoras's Theorem]] ($q$ and $P q$ are orthogonal)
+}}
+{{eqn | r = c^2 + 1
+      | c = since $\norm {P q} = c$ and $\norm q = 1$.
+}}
+{{end-eqn}}
+Now
+{{begin-eqn}}
+{{eqn | l = \norm {P \frac y {\norm y} }^2
+      | r = \frac 1 {\norm y^2} \norm {P \paren {c q + \frac 1 c P q} }^2
+}}
+{{eqn | r = \frac 1 {c^2 + 1} \norm {c P q + \frac 1 c P^2 q}^2
+}}
+{{eqn | r = \frac 1 {c^2 + 1} \norm {P q \paren {c + \frac 1 c} }^2
+      | c = $P$ is [[Idempotent Operators|idempotent]]
+}}
+{{eqn | r = \frac {\norm {P q}^2} {c^2 + 1} \paren {c + \frac 1 c}^2
+}}
+{{eqn | r = \frac {c^2 \paren {c + \frac 1 c}^2} {c^2 + 1}
+}}
+{{eqn | r = c^2 + 1
+}}
+{{eqn | o = >
+      | r = 1
+}}
+{{end-eqn}}
+{{qed}}
+mdztc7gcmgwuwht6tbkijp9rc08rkpb
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 3 y' - 10 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 3 y' - 10 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{2 x} + C_2 e^{-5 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + 3 m - 10 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 2$
+:$m_2 = -5$
+These are [[Definition:Real Number|real]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{2 x} + C_2 e^{-5 x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+a1l28sxz22kzbjvcnihq84ksi3eenj6
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 3 y' - 10 y = 6 exp 4 x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 3 y' - 10 y = 6 e^{4 x}$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{2 x} + C_2 e^{-5 x} + \dfrac {e^{4 x} } 3$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 3$
+:$q = -10$
+:$\map R x = 6 e^{4 x}$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' + 3 y' - 10 y = 0$
+From [[Linear Second Order ODE/y'' + 3 y' - 10 y = 0|Linear Second Order ODE: $y'' + 3 y' - 10 y = 0$]], this has the [[Definition:General Solution|general solution]]:
+:$y_g = C_1 e^{2 x} + C_2 e^{-5 x}$
+We have that:
+:$\map R x = 6 e^{4 x}$
+and so from the [[Method of Undetermined Coefficients/Exponential|Method of Undetermined Coefficients for the Exponential function]]:
+:$y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$
+where:
+:$K = 6$
+:$a = 4$
+:$p = 3$
+:$q = -10$
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \dfrac {6 e^{4 x} } {4^2 + 3 \times 4 - 10}
+      | c = 
+}}
+{{eqn | r = \dfrac {6 e^{4 x} } {16 + 12 - 10}
+      | c = 
+}}
+{{eqn | r = \dfrac {e^{4 x} } 3
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 e^{2 x} + C_2 e^{-5 x} + \dfrac {e^{4 x} } 3$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 4 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs, Linear Second Order ODE: y'' + 4 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 4 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \cos 2 x + C_2 \sin 2 x$
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 4 y = 3 sine x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 4 y = 3 \sin x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = 4$
+:$\map R x = 3 \sin x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' + 4 y = 0$
+From [[Linear Second Order ODE/y'' + 4 y = 0|Linear Second Order ODE: $y'' + 4 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 \cos 2 x + C_2 \sin 2 x$
+We have that:
+:$\map R x = 3 \sin x$
+and it is noted that $\sin x$ is not itself a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Sine and Cosine|Method of Undetermined Coefficients for the Sine and Cosine functions]]:
+:$y_p = A \sin x + B \cos x$
+for some $A$ and $B$ to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A \sin x + B \cos x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = A \cos x = B \sin x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = -A \sin x - B \cos x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = -A \sin x - B \cos x + 4 \paren {A \sin x + B \cos x}
+      | r = 3 \sin x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {-A + 4 A} \sin x
+      | r = 3 \sin x
+      | c = 
+}}
+{{eqn | l = -B \cos x + B \cos x
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 3 A
+      | r = 3
+      | c = 
+}}
+{{eqn | l = -B \cos x + 4 B \cos x
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | r = 1
+      | c = 
+}}
+{{eqn | l = B
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = y_p
+      | r = \sin x
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 10 y' + 25 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 10 y' + 25 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{-5 x} + C_2 x e^{-5 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + 10 m + 25 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = m_2 = -5$
+These are [[Definition:Real Number|real]] and equal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{-5 x} + C_2 x e^{-5 x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 10 y' + 25 y = 14 exp -5 x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 10 y' + 25 y = 14 e^{-5 x}$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 10$
+:$q = 25$
+:$\map R x = 14 e^{-5 x}$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' + 10 y' + 25 y = 0$
+From [[Linear Second Order ODE/y'' + 10 y' + 25 y = 0|Linear Second Order ODE: $y'' + 10 y' + 25 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 e^{-5 x} + C_2 x e^{-5 x}$
+We have that:
+:$\map R x = 14 e^{-5 x}$
+and it is noted that $14 e^{-5 x}$ is a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Exponential|Method of Undetermined Coefficients for the Exponential function]]:
+:$y_p = A x^2 e^{-5 x}$
+where:
+:$A = \dfrac {14} 2$
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 e^{-5 x} + C_2 x e^{-5 x} + 7 x^2 e^{-5 x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 2 y' + 5 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 2 y' + 5 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 - 2 m + 5 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 1 + 2 i$
+:$m_2 = 1 - 2 i$
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+jzicno37hdhhn46n8hu6kaerroomr72
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 2 y' + 5 y = 25 x^2 + 12}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 2 y' + 5 y = 25 x^2 + 12$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x} + 2 + 4 x + 5 x^2$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = -2$
+:$q = 5$
+:$\map R x = 25 x^2 + 12$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' - 2 y' + 5 y = 0$
+From [[Linear Second Order ODE/y'' - 2 y' + 5 y = 0|Linear Second Order ODE: $y'' - 2 y' + 5 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
+We have that:
+:$\map R x = 25 x^2 + 12$
+and it is noted that $25 x^2 + 12$ is not itself a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Polynomial|Method of Undetermined Coefficients for Polynomials]]:
+:$y_p = A_0 + A_1 x + A_2 x^2$
+for $A_n$ to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A_0 + A_1 x + A_2 x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = A_1 + 2 A_2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = 2 A_2
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = 2 A_2 - 2 \paren {A_1 + 2 A_2 x} + 5 \paren {A_0 + A_1 x + A_2 x^2}
+      | r = 25 x^2 + 12
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 5 A_2 x^2
+      | r = 25 x^2
+      | c = equating powers
+}}
+{{eqn | l = -4 A_2 x + 5 A_1 x
+      | r = 0
+      | c = 
+}}
+{{eqn | l = 2 A_2 - 2 A_1 + 5 A_0
+      | r = 12
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A_2
+      | r = 5
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -4 \times 5 x + 5 A_1 x
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -4 + A_1
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A_1
+      | r = 4
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 2 \times 5 - 2 \times 4 + 5 A_0
+      | r = 12
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 10 - 8 + 5 A_0
+      | r = 12
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 5 A_0
+      | r = 10
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A_0
+      | r = 2
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x} + 2 + 4 x + 5 x^2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + y' - 12 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + y' - 12 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{3 x} + C_2 e^{-4 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + m - 12 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 3$
+:$m_2 = -4$
+These are [[Definition:Real Number|real]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{3 x} + C_2 e^{-4 x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+rwwnplqveq38970wxbsx99ciiyn8p2t
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 2 y' + 10 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 2 y' + 10 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^{-x} \paren {C_1 \cos 3 x + C_2 \sin 3 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + 2 m + 10 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = -1 + 3 i$
+:$m_2 = -1 - 3 i$
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = e^{-x} \paren {C_1 \cos 3 x + C_2 \sin 3 x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+pd5h9hruyy2du3ixzw31q3zyp2m3zl4
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 4 y' + 4 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 4 y' + 4 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = \paren {C_1 + C_2 x} e^{-2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + 4 m + 4 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = m_2 = -2$
+These are [[Definition:Real Number|real]] and equal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{-2 x} + C_2 x e^{-2 x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + y' - 2 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + y' - 2 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^x + C_2 e^{-2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + m - 2 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 1$
+:$m_2 = -2$
+These are [[Definition:Real Number|real]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^x + C_2 e^{-2 x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+kjp37andmikkls9xbinool6klbb8a0j
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 4 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs, Linear Second Order ODE: y'' - 4 y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 4 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{2 x} + C_2 e^{-2 x}$
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/y'' - y' - 6 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - y' - 6 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{3 x} + C_2 e^{-2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 - m - 6 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 3$
+:$m_2 = -2$
+These are [[Definition:Real Number|real]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{3 x} + C_2 e^{-2 x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+q6aiorh940jz104l1655sf7955s36e4
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - y' - 6 y = 20 exp -2 x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - y' - 6 y = 20 e^{-2 x}$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = -1$
+:$q = -6$
+:$\map R x = 20 e^{-2 x}$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' - y' - 6 y = 0$
+From [[Linear Second Order ODE/y'' - y' - 6 y = 0|Linear Second Order ODE: $y'' - y' - 6 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$
+We have that:
+:$\map R x = 20 e^{-2 x}$
+and so from the [[Method of Undetermined Coefficients/Exponential|Method of Undetermined Coefficients for the Exponential function]]:
+:$y_p = \dfrac {K x e^{a x} } {2 a + p}$
+where:
+:$K = 20$
+:$a = -2$
+:$p = -1$
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \dfrac {20 x e^{-2 x} } {2 \times \paren {-2} - 1}
+      | c = 
+}}
+{{eqn | r = \dfrac {20 x e^{-2 x} } {-5}
+      | c = 
+}}
+{{eqn | r = -4 x e^{-2 x}
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 3 y' + 2 y = 14 sine 2 x - 18 cosine 2 x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 3 y' + 2 y = 14 \sin 2 x - 18 \cos 2 x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = -3$
+:$q = 2$
+:$\map R x = 14 \sin 2 x - 18 \cos 2 x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' - 3 y' + 2 y = 0$
+From [[Linear Second Order ODE/y'' - 3 y' + 2 y = 0|Linear Second Order ODE: $y'' - 3 y' + 2 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 e^x + C_2 e^{2 x}$
+We have that:
+:$\map R x = 14 \sin 2 x - 18 \cos 2 x$
+and it is noted that $14 \sin 2 x - 18 \cos 2 x$ is not itself a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Sine and Cosine|Method of Undetermined Coefficients for Sine and Cosine]]:
+:$y_p = A \sin 2 x + B \cos 2 x$
+where $A$ and $B$ are to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A \sin 2 x + B \cos 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = 2 A \cos 2 x - 2 B \sin 2 x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = -4 A \sin 2 x - 4 B \cos 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = -4 A \sin 2 x - 4 B \cos 2 x - 3 \paren {2 A \cos 2 x - 2 B \sin 2 x} + 2 \paren {A \sin 2 x + B \cos 2 x}
+      | r = 14 \sin 2 x - 18 \cos 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -4 A \sin 2 x + 6 B \sin 2 x + 2 A \sin 2 x
+      | r = 14 \sin 2 x
+      | c = equating coefficients
+}}
+{{eqn | l = -4 B \cos 2 x - 6 A \cos 2 x + 2 B \cos 2 x
+      | r = - 18 \cos 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -2 A + 6 B
+      | r = 14
+      | c = 
+}}
+{{eqn | l = -2 B - 6 A
+      | r = -18
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -A + 3 B
+      | r = 7
+      | c = 
+}}
+{{eqn | l = 3 A + B
+      | r = 9
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -A + 3 \paren {-3 A + 9}
+      | r = 7
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -10 A
+      | r = -20
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | r = 2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -2 + 3 B
+      | r = 7
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = B
+      | r = 3
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + 2 \sin 2 x + 3 \cos 2 x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + y = 2 cosine x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + y = 2 \cos x$
+has the [[Definition:General Solution|general solution]]:
+:$y = C_1 \sin x + C_2 \cos x + x \cos x$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = 1$
+:$\map R x = 2 \cos x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' + y = 0$
+From [[Linear Second Order ODE/y'' + y = 0|Linear Second Order ODE: $y'' + y = 0$]], this has the [[Definition:General Solution|general solution]]:
+:$y_g = C_1 \sin x + C_2 \cos x$
+We have that:
+:$\map R x = 2 \cos x$
+and it is noted that $2 \cos x$ is a [[Definition:Particular Solution|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Sine and Cosine|Method of Undetermined Coefficients for Sine and Cosine]]:
+:$y_p = A x \sin x + B x \cos x$
+where $A$ and $B$ are to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A x \sin x + B x \cos x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = A x \cos x - B x \sin x + A \sin x + B \cos x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = -A x \sin x - B x \cos x + A \cos x - B \sin x + A \cos x - B \sin x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | r = -A x \sin x - B x \cos x + 2 A \cos x - 2 B \sin x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = -A x \sin x - B x \cos x + 2 A \cos x - 2 B \sin x + A x \sin x + B x \cos x
+      | r = 2 \cos x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -2 B \sin x
+      | r = 0
+      | c = equating coefficients
+}}
+{{eqn | l = 2 A \cos x
+      | r = 2 \cos x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = B
+      | r = 0
+      | c = 
+}}
+{{eqn | l = A
+      | r = 1
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \sin x + C_2 \cos x + x \sin x$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 2 y' = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 2 y' = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 + C_2 e^{2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 - 2 m = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 0$
+:$m_2 = 2$
+These are [[Definition:Real Number|real]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{0 x} + C_2 e^{2 x} = C_1 + C_2 e^{2 x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+inhed7gc83phylrmx05mo5zksgxz5cc
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 2 y' = 12 x - 10}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 2 y' = 12 x - 10$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 + C_2 e^{2 x} + 2 x - 3 x^2$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 2$
+:$q = 0$
+:$\map R x = 12 x - 10$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' - 2 y' = 0$
+From [[Linear Second Order ODE/y'' - 2 y' = 0|Linear Second Order ODE: $y'' - 2 y' = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 + C_2 e^{2 x}$
+We have that:
+:$\map R x = 12 x - 10$
+and it is noted that $12 x - 10$ is not itself a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Polynomial|Method of Undetermined Coefficients for Polynomials]]:
+:$y_p = A_0 + A_1 x + A_2 x^2$
+where $A$ and $B$ are to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A_0 + A_1 x + A_2 x^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = A_1 + 2 A_2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = 2 A_2
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = 2 A_2 - 2 \paren {A_1 + 2 A_2 x}
+      | r = 12 x - 10
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -4 A_2 x
+      | r = 12 x
+      | c = equating coefficients
+}}
+{{eqn | l = 2 A_2 - 2 A_1
+      | r = -10
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A_2
+      | r = -3
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -6 - 2 A_1
+      | r = -10
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -2 A_1
+      | r = -4
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A_1
+      | r = 2
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 + C_2 e^{2 x} + 2 x - 3 x^2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Directed iff Finite Subsets have Upper Bounds}
+Tags: Preorder Theory
+
+\begin{theorem}
+Let $\left({S, \precsim}\right)$ be a [[Definition:Preordered Set|preordered set]].
+Let $H$ be a [[Definition:Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Then $H$ is [[Definition:Directed Subset|directed]] {{iff}}
+:for every a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $H$
+::$\exists h \in H: \forall a \in A: a \precsim h$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $R$ be [[Definition:Directed Subset|directed]].
+We will prove by [[Principle of Mathematical Induction|induction]] of the [[Definition:Cardinality|cardinality]] of [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $H$.
+==== Base case ====
+:$\forall A \subseteq H: \left\vert{A}\right\vert = 0 \implies \exists h \in H: \forall a \in A: a \precsim h$
+where $\left\vert{A}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $A$.
+Let $A \subseteq H$ such that
+:$\left\vert{A}\right\vert = 0$
+By [[Cardinality of Empty Set]]:
+:$A = \varnothing$
+By definition of [[Definition:Empty Set|empty set]]:
+:$\exists h: h \in H$
+Thus by definition of [[Definition:Empty Set|empty set]]:
+:$\exists h \in H: \forall a \in A: a \precsim h$
+==== Induction Hypothesis  ====
+:$\forall A \subseteq H: \left\vert{A}\right\vert = n \implies \exists h \in H: \forall a \in A: a \precsim h$
+==== Induction Step ====
+:$\forall A \subseteq H: \left\vert{A}\right\vert = n+1 \implies \exists h \in H: \forall a \in A: a \precsim h$
+Let $A \subseteq H$ such that
+:$\left\vert{A}\right\vert = n+1$
+By definition of [[Definition:Cardinality of Finite Set|cardinality of finite set]]:
+:$A \sim \N_{< n+1}$
+where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]].
+By definition of [[Definition:Set Equivalence|set equivalence]]:
+:there exists a [[Definition:Bijection|bijection]] $f: \N_{< n+1} \to A$
+By [[Restriction of Injection is Injection]]:
+:$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is an [[Definition:Injection|injection]].
+By definition
+:$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is a [[Definition:Surjection|surjection]].
+By definition
+:$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is a [[Definition:Bijection|bijection]].
+By definition of [[Definition:Set Equivalence|set equivalence]]:
+:$\N_{< n} \sim f^\to\left({\N_{< n}}\right)$
+By definition of [[Definition:Cardinality of Finite Set|cardinality of finite set]]:
+:$\left\vert{f^\to\left({\N_{< n}}\right)}\right\vert = n$
+By definitions of [[Definition:Image of Subset under Mapping|image of set]] and [[Definition:Subset|subset]]:
+:$f^\to\left({\N_{< n}}\right) \subseteq A$
+By [[Subset Relation is Transitive]]:
+:$f^\to\left({\N_{< n}}\right) \subseteq H$
+By Induction Hypothesis:
+:$\exists h \in H: \forall a \in f^\to\left({\N_{< n}}\right): a \precsim h$
+By definition $\N_{< n+1}$
+:$n \in \N_{< n+1}$
+By definition of [[Definition:Mapping|mapping]]:
+:$f \left({n}\right) \in A$
+By definition of [[Definition:Subset|subset]]:
+:$f \left({n}\right) \in H$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists h' \in H: f \left({n}\right) \precsim h' \land  h \precsim h'$
+{{begin-eqn}}
+{{eqn |l = A
+      |r = f^\to\left({\N_{< n+1} }\right)
+| c = definition of [[Definition:Surjection|surjection]]
+}}
+{{eqn |r =  f^\to\left({\N_{< n}\cup \left\{ {n}\right\} }\right)
+| c = definition of [[Definition:Initial Segment of Natural Numbers|initial segment of natural numbers]]
+}}
+{{eqn |r = f^\to\left({\N_{< n} }\right) \cup f^\to\left({\left\{ {n}\right\} }\right)
+| c =[[Image of Union under Mapping]]
+}}
+{{eqn |r = f^\to\left({\N_{< n} }\right) \cup \left\{ {f\left({n }\right)}\right\}
+| c =[[Image of Singleton under Mapping]]
+}}
+{{end-eqn}}
+Let $a \in A$.
+Then by definitions of [[Definition:Set Union|union]] and [[Definition:Singleton|singleton]]:
+:$a \in f^\to\left({\N_{< n} }\right) \lor a = f\left({n }\right)$
+So:
+:$a \precsim h \lor a = f\left({n }\right)$
+Thus by definition of [[Definition:Transitivity|transitivity]]:
+:$a \precsim h'$
+Thus:
+:$\exists h' \in H: \forall a \in A: a \precsim h'$
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:for every a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $H$
+::$\exists h \in H: \forall a \in A: a \precsim h$
+Let $a, b \in H$
+By definition of [[Definition:Subset|subset]]:
+:$\left\{ {a, b}\right\} \subseteq H$
+:$\left\{ {a, b}\right\}$ is [[Definition:Finite Set|finite]]
+By assumption:
+:$\exists h \in H: \forall c \in \left\{ {a, b}\right\}: c \precsim h$
+Thus by definition of [[Definition:Unordered Tuple|unordered tuple]]:
+:$\exists h \in H: a \precsim h \land b \precsim h$
+Thus by definition
+:$H$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + k^2 y = sine b x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + k^2 y = \sin b x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = \begin{cases} C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2} & : b \ne k \\
+C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k} & : b = k 
+\end{cases}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = k^2$
+:$\map R x = \sin b x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' + k^2 y = 0$
+From [[Linear Second Order ODE/y'' + k^2 y = 0|Linear Second Order ODE: $y'' + k^2y = 0$]], this has the [[Definition:General Solution|general solution]]:
+:$y_g = C_1 \sin k x + C_2 \cos k x$
+We have that:
+:$\map R x = \sin b x$
+There are two cases to address:
+:$b = k$
+:$b \ne k$
+First suppose that $b = k$.
+It is noted that $\sin b x = \sin k x$ is a [[Definition:Particular Solution|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Sine and Cosine|Method of Undetermined Coefficients for Sine and Cosine]]:
+:$y_p = A x \sin k x + B x \cos k x$
+where $A$ and $B$ are to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A x \sin k x + B x \cos k x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = A k x \cos k x - B k x \sin k x + A \sin k x + B \cos k x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]], [[Product Rule]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = -A k^2 x \sin k x - B k^2 x \cos k x + A k \cos k x - B k \sin k x + A k \cos k x - B k \sin k x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]], [[Product Rule]]
+}}
+{{eqn | r = -A k^2 x \sin k x - B k^2 x \cos k x + 2 A k \cos k x - 2 B k \sin k x
+      | c = 
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = -A k^2 x \sin k x - B k^2 x \cos k x + 2 A k \cos k x - 2 B k \sin k x + k^2 \paren {A x \sin k x + B x \cos k x}
+      | r = \sin k x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A \paren {k^2 - k^2} x \sin k x - 2 B k \sin k x
+      | r = \sin k x
+      | c = equating coefficients
+}}
+{{eqn | l = B \paren {k^2 - k^2} x \cos k x + 2 A k \cos k x
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = - 2 B k
+      | r = 1
+      | c = 
+}}
+{{eqn | l = 2 A k
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = B
+      | r = -\frac 1 {2 k}
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k}$
+{{qed|lemma}}
+Now suppose that $b \ne k$.
+It is noted that $\sin b x$ is not a [[Definition:Particular Solution|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Sine and Cosine|Method of Undetermined Coefficients for Sine and Cosine]]:
+:$y_p = A \sin b x + B \cos b x$
+where $A$ and $B$ are to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A \sin b x + B \cos b x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = A b \cos b x - B b \sin b x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = -A b^2 \sin b x - B b^2 \cos b x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = -A b^2 \sin b x - B b^2 \cos b x + k \paren {A \sin b x + B \cos b x}
+      | r = \sin b x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A \paren {k^2 - b^2} \sin b x
+      | r = \sin b x
+      | c = equating coefficients
+}}
+{{eqn | l = B \paren {k^2 - b^2} \cos b x
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | r = 0
+      | c = 
+}}
+{{eqn | l = B
+      | r = \frac 1 {k^2 - b^2}
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Combination of Solutions to Non-Homogeneous LSOODE with same Homogeneous Part}
+Tags: Linear Second Order ODEs
+
+\begin{theorem}
+Let $\map {y_1} x$ be a [[Definition:Particular Solution|particular solution]] of the [[Definition:Linear Second Order ODE|linear second order ODE]]:
+:$(1): \quad y'' + \map P x y' + \map Q x y = \map {R_1} x$
+Let $\map {y_2} x$ be a [[Definition:Particular Solution|particular solution]] of the [[Definition:Linear Second Order ODE|linear second order ODE]]:
+:$(2): \quad y'' + \map P x y' + \map Q x y = \map {R_2} x$
+Then $\map y x = \map {y_1} x + \map {y_2} x$ is a [[Definition:Particular Solution|particular solution]] of the [[Definition:Linear Second Order ODE|linear second order ODE]]:
+:$(3): \quad y'' + \map P x y' + \map Q x y = \map {R_1} x + \map {R_2} x$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | n = 4
+      | l = {y_1}'' + \map P x {y_1}' + \map Q x y_1
+      | r = \map {R_1} x
+      | c = as $y_1$ is a [[Definition:Particular Solution|particular solution]] to $(1)$
+}}
+{{eqn | n = 5
+      | l = {y_2}'' + \map P x {y_2}' + \map Q x y_2
+      | r = \map {R_2} x
+      | c = as $y_2$ is a [[Definition:Particular Solution|particular solution]] to $(2)$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren { {y_1}'' + {y_2}''} + \map P x \paren { {y_1}' + {y_2}'} + \map Q x \paren {y_1 + y_2}
+      | r = \map {R_1} x + \map {R_2} x
+      | c = adding $(4)$ and $(5)$
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {y_1 + y_2}'' + \map P x \paren {y_1 + y_2}' + \map Q x \paren {y_1 + y_2}
+      | r = \map {R_1} x + \map {R_2} x
+      | c = [[Linear Combination of Derivatives]]
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x + 6 cosine x + 8 x^2 - 4 x}
+Tags: Examples of Constant Coefficient LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 4 y = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x + 2 \cos x - 1 - x + 2 x^2$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = 4$
+:$\map R x = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' + 4 y = 0$
+From [[Linear Second Order ODE/y'' + 4 y = 0|Linear Second Order ODE: $y'' + 4 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 \sin 2 x + C_2 \cos 2 x$
+We have that:
+:$\map R x = 4 \cos 2 x + 6 \cos x + 8 x^2 - 4 x$
+Let:
+:$\map R x = \map {R_1} x + \map {R_2} x + \map {R_3} x$
+where:
+:$\map {R_1} x = 4 \cos 2 x$
+:$\map {R_2} x = 6 \cos x$
+:$\map {R_3} x = 8 x^2 - 4 x$
+Consider in turn the solutions to:
+:$y'' + p y' + q y = \map {R_1} x$
+From [[Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x|Linear Second Order ODE: $y'' + 4 y = 4 \cos 2 x$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x$
+:$y'' + p y' + q y = \map {R_2} x$
+From [[Linear Second Order ODE/y'' + 4 y = 6 cosine x|Linear Second Order ODE: $y'' + 4 y = 6 \cos x$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin 2 x + C_2 \cos 2 x + 2 \cos x$
+:$y'' + p y' + q y = \map {R_3} x$
+From [[Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x|Linear Second Order ODE: $y'' + 4 y = 8 x^2 - 4 x$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$
+It follows from [[Combination of Solutions to Non-Homogeneous LSOODE with same Homogeneous Part]] that the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$ is:
+:$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x + 2 \cos x - 1 - x + 2 x^2$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 4 y = 4 \cos 2 x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = 4$
+:$\map R x = 4 \cos 2 x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' + 4 y = 0$
+From [[Linear Second Order ODE/y'' + 4 y = 0|Linear Second Order ODE: $y'' + 4 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 \sin 2 x + C_2 \cos 2 x$
+It is noted that $4 \cos 2 x$ is a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Sine and Cosine|Method of Undetermined Coefficients for Sine and Cosine]]:
+:$y_p = A x \sin 2 x + B x \cos 2 x$
+where $A$ and $B$ are to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A x \sin 2 x + B x \cos 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = 2 A x \cos 2 x - 2 B x \sin 2 x + A \sin 2 x + B \cos 2 x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]], [[Product Rule]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = -4 A x \sin 2 x - 4 B x \cos 2 x + 2 A \cos 2 x - 2 B \sin 2 x + 2 A \cos 2 x - 2 B \sin 2 x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]], [[Product Rule]]
+}}
+{{eqn | r = -4 A x \sin 2 x - 4 B x \cos 2 x + 4 A \cos 2 x - 4 B \sin 2 x
+      | c = 
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = -4 A x \sin 2 x - 4 B x \cos 2 x + 4 A \cos 2 x - 4 B \sin 2 x + 4 \paren {A x \sin 2 x + B x \cos 2 x}
+      | r = 4 \cos 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A \paren {4 - 4} x \sin 2 x - 4 B \sin 2 x
+      | r = 0
+      | c = equating coefficients
+}}
+{{eqn | l = B \paren {4 - 4} x \cos 2 x + 4 A \cos 2 x
+      | r = 4 \cos 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = - 4 B
+      | r = 0
+      | c = 
+}}
+{{eqn | l = 4 A
+      | r = 4
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | r = 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = B
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + x \sin 2 x$
+is the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$.
+{{qed}}
+[[Category:Examples of Constant Coefficient LSOODEs]]
+[[Category:Examples of Method of Undetermined Coefficients]]
+ckxfh56cvg4csyr4ue05k2mmlea9987
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 4 y = 6 cosine x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 4 y = 6 \cos x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin 2 x + C_2 \cos 2 x + 2 \cos x$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = 4$
+:$\map R x = 6 \cos x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' + 4 y = 0$
+From [[Linear Second Order ODE/y'' + 4 y = 0|Linear Second Order ODE: $y'' + 4 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 \sin 2 x + C_2 \cos 2 x$
+It is noted that $\cos x$ is not a [[Definition:Particular Solution to Differential Equation|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Sine and Cosine|Method of Undetermined Coefficients for Sine and Cosine]]:
+:$y_p = A \sin x + B \cos x$
+where $A$ and $B$ are to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A \sin x + B \cos x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = A \cos x - B \sin x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = -A \sin x - B \cos x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = -A \sin x - B \cos x + 4 \paren {A \sin x + B \cos x}
+      | r = 6 \cos x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 3 A \sin x
+      | r = 0
+      | c = equating coefficients
+}}
+{{eqn | l = 3 B \cos x
+      | r = 6 \cos x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | r = 0
+      | c = 
+}}
+{{eqn | l = B
+      | r = 2
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + 2 \cos x$
+is the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$.
+{{qed}}
+[[Category:Examples of Constant Coefficient LSOODEs]]
+[[Category:Examples of Method of Undetermined Coefficients]]
+i707s896lgr763ay1nl3jlne0l5pbdl
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 4 y = 8 x^2 - 4 x}
+Tags: Examples of Constant Coefficient LSOODEs, Linear Second Order ODE: y'' + 4 y = 8 x^2 - 4 x
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 4 y = 8 x^2 - 4 x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/y'' + y = cosecant x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Variation of Parameters
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + y = \csc x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin x + C_2 \cos x - x \cos x + \sin x \map \ln {\sin x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $\paren 1$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = 1$
+:$\map R x = \csc x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$\paren 2: \quad y'' + y = 0$
+From [[Linear Second Order ODE/y'' + y = 0|Linear Second Order ODE: $y'' + y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 \sin x + C_2 \cos x$
+It remains to find a [[Definition:Particular Solution|particular solution]] $y_p$ to $\paren 1$.
+Expressing $y_g$ in the form:
+:$y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$
+we have:
+{{begin-eqn}}
+{{eqn | l = \map {y_1} x
+      | r = \sin x
+      | c = 
+}}
+{{eqn | l = \map {y_2} x
+      | r = \cos x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map {y_1'} x
+      | r = \cos x
+      | c = [[Derivative of Sine Function]]
+}}
+{{eqn | l = \map {y_2'} x
+      | r = -\sin x
+      | c = [[Derivative of Cosine Function]]
+}}
+{{end-eqn}}
+By the [[Method of Variation of Parameters]], we have that:
+:$y_p = v_1 y_1 + v_2 y_2$
+where:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = -\int \frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+where $\map W {y_1, y_2}$ is the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = {{Defof|Wronskian}}
+}}
+{{eqn | r = \sin x \paren {- \sin x} - \cos x \cos x
+      | c = 
+}}
+{{eqn | r = -\paren {\sin^2 x + \cos^2 x}
+      | c = 
+}}
+{{eqn | r = -1
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = -\int \frac {\cos x \csc x} {-1} \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {\cos x} {\sin x} \rd x
+      | c = {{Defof|Cosecant}}
+}}
+{{eqn | r = \map \ln {\sin x}
+      | c = [[Primitive of Cotangent Function|Primitive of $\cot x$]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {\sin x \csc x} {-1} \rd x
+      | c = 
+}}
+{{eqn | r = -\int \frac {\sin x} {\sin x} \rd x
+      | c = {{Defof|Cosecant}}
+}}
+{{eqn | r = -x
+      | c = 
+}}
+{{end-eqn}}
+It follows that:
+:$y_p = \sin x \map \ln {\sin x} - x \cos x$
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \sin x + C_2 \cos x + \sin x \map \ln {\sin x} - x \cos x$
+is the [[Definition:General Solution to Differential Equation|general solution]] to $\paren 1$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 2 y' + y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 2 y' + y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^x + C_2 x e^x$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 - 2 m + 1 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = m_2 = 1$
+These are [[Definition:Real Number|real]] and equal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^x + C_2 x e^x$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+c8k5v3yna87ldu4rn2mriauu6cb43cu
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 2 y' + y = 2 x}
+Tags: Examples of Constant Coefficient LSOODEs, Linear Second Order ODE: y'' - 2 y' + y = 2 x
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 2 y' + y = 2 x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^x + C_2 x e^x + 2 x + 4$
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/y'' - y' - 6 y = exp -x}
+Tags: Examples of Constant Coefficient LSOODEs, Linear Second Order ODE: y'' - y' - 6 y = exp -x
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - y' - 6 y = e^{-x}$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$
+\end{theorem}<|endoftext|>
+\section{Filtered iff Finite Subsets have Lower Bounds}
+Tags: Preorder Theory
+
+\begin{theorem}
+Let $\struct {S, \precsim}$ be a [[Definition:Preordered Set|preordered set]].
+Let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Then $H$ is [[Definition:Filtered Subset|filtered]] {{iff}}:
+:for every a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $H$
+::$\exists h \in H: \forall a \in A: h \precsim a$
+\end{theorem}
+
+\begin{proof}
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]] of the proof of [[Directed iff Finite Subsets have Upper Bounds]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Tangent of Half Angle plus Quarter Pi}
+Tags: Trigonometric Identities, Tangent Function, Secant Function
+
+\begin{theorem}
+:$\map \tan {\dfrac x 2 + \dfrac \pi 4} = \tan x + \sec x$
+\end{theorem}
+
+\begin{proof}
+Firstly, we have:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \tan x
+      | r = \frac {2 \tan \frac x 2} {1 - \tan ^2 \frac x 2}
+      | c = [[Double Angle Formula for Tangent]]
+}}
+{{end-eqn}}
+Then:
+{{begin-eqn}}
+{{eqn | l = \map \tan {\frac x 2 + \frac \pi 4}
+      | r = \frac {\tan \frac x 2 + \tan \frac \pi 4} {1 - \tan \frac x 2 \tan \frac \pi 4}
+      | c = [[Tangent of Sum]]
+}}
+{{eqn | r = \frac {\tan \frac x 2 + 1} {1 - \tan \frac x 2}
+      | c = [[Tangent of 45 Degrees|Tangent of $45 \degrees$]]
+}}
+{{eqn | r = \frac {\paren {\tan \frac x 2 + 1} \paren {\tan \frac x 2 + 1} } {\paren {1 - \tan \frac x 2} \paren {\tan \frac x 2 + 1} }
+}}
+{{eqn | r = \frac {\tan^2 \frac x 2 + 2 \tan \frac x 2 + 1} {1 - \tan^2 \frac x 2}
+      | c = [[Difference of Two Squares]], [[Square of Sum]]
+}}
+{{eqn | r = \frac {2 \tan \frac x 2} {1 - \tan^2 \frac x 2} + \frac {\tan^2 \frac x 2 + 1} {1 - \tan^2 \frac x 2}
+}}
+{{eqn | r = \tan x + \frac {\tan^2 \frac x 2 + 1} {1 - \tan^2 \frac x 2}
+      | c = [[Double Angle Formula for Tangent]]: see $(1)$ above
+}}
+{{eqn | r = \tan x + \frac {\sin^2 \frac x 2 + \cos^2 \frac x 2} {\cos^2 \frac x 2 - \sin^2 \frac x 2}
+      | c = multiplying [[Definition:Denominator|Denominator]] and [[Definition:Numerator|Numerator]] by $\cos^2 \frac x 2$
+}}
+{{eqn | r = \tan x + \frac {\sin^2 \frac x 2 + \cos^2 \frac x 2} {\cos 2 \frac x 2}
+      | c = [[Double Angle Formula for Cosine]]
+}}
+{{eqn | r = \tan x + \frac 1 {\cos x}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{eqn | r = \tan x + \sec x
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Trigonometric Identities]]
+[[Category:Tangent Function]]
+[[Category:Secant Function]]
+0wwp1p2ickye49m0s6kfoq8lv5b8409
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 4 y = tangent 2 x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Variation of Parameters
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 4 y = \tan 2 x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \cos 2 x + C_2 \sin 2 x - \dfrac 1 4 \cos 2 x \, \map \ln {\sec 2 x + \tan 2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = 4$
+:$\map R x = \tan 2 x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' + 4 y = 0$
+From [[Linear Second Order ODE/y'' + 4 y = 0|Linear Second Order ODE: $y'' + 4 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 \cos 2 x + C_2 \sin 2 x$
+It remains to find a [[Definition:Particular Solution to Differential Equation|particular solution]] $y_p$ to $(1)$.
+Expressing $y_g$ in the form:
+:$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$
+we have:
+{{begin-eqn}}
+{{eqn | l = \map {y_1} x
+      | r = \cos 2 x
+      | c = 
+}}
+{{eqn | l = \map {y_2} x
+      | r = \sin 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map { {y_1}'} x
+      | r = -2 \sin 2 x
+      | c = [[Derivative of Sine Function]]
+}}
+{{eqn | l = \map { {y_2}'} x
+      | r = 2 \cos 2 x
+      | c = [[Derivative of Cosine Function]]
+}}
+{{end-eqn}}
+By the [[Method of Variation of Parameters]], we have that:
+:$y_p = v_1 y_1 + v_2 y_2$
+where:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+where $\map W {y_1, y_2}$ is the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = {{Defof|Wronskian}}
+}}
+{{eqn | r = \cos 2 x \paren {2 \cos 2 x} - \sin 2 x \paren {-2 \sin 2 x}
+      | c = 
+}}
+{{eqn | r = 2 \paren {\cos^2 2 x + \sin^2 2 x}
+      | c = 
+}}
+{{eqn | r = 2
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int -\frac {\sin 2 x \tan 2 x} 2 \rd x
+      | c = 
+}}
+{{eqn | r = -\frac 1 2 \int \frac {\sin^2 2 x} {\cos 2 x} \rd x
+      | c = 
+}}
+{{eqn | r = -\frac 1 2 \paren {-\frac {\sin 2 x} 2 + \frac 1 2 \ln \map \tan {x + \frac \pi 4} }
+      | c = [[Primitive of Square of Sine of a x over Cosine of a x|Primitive of $\dfrac {\sin^2 a x} {\cos a x}$]]
+}}
+{{eqn | r = \frac 1 4 \paren {\sin 2 x - \map \ln {\sec 2 x + \tan 2 x} }
+      | c = [[Tangent of Half Angle plus Quarter Pi|Tangent of Half Angle plus $\dfrac \pi 4$]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {\cos 2 x \tan 2 x} 2 \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \int \sin 2 x \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \paren {-\frac {\cos 2 x} 2}
+      | c = [[Primitive of Sine of a x|Primitive of $\sin a x$]]
+}}
+{{eqn | r = -\frac 1 4 \cos 2 x
+      | c = 
+}}
+{{end-eqn}}
+It follows that:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \frac 1 4 \paren {\sin 2 x - \map \ln {\sec 2 x + \tan 2 x} } \cos 2 x - \frac 1 4 \cos 2 x \sin 2 x
+      | c = 
+}}
+{{eqn | r = -\frac 1 4 \cos 2 x \, \map \ln {\sec 2 x + \tan 2 x}
+      | c = simplifying
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \cos 2 x + C_2 \sin 2 x - \dfrac 1 4 \cos 2 x \, \map \ln {\sec 2 x + \tan 2 x}$
+is the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 2 y' + y = exp -x log x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Variation of Parameters
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 2 y' + y = e^{-x} \ln x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{-x} + C_2 x e^{-x} - \dfrac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 2$
+:$q = 1$
+:$\map R x = e^{-x} \ln x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' + 2 y' + y = 0$
+From [[Linear Second Order ODE/y'' + 2 y' + y = 0|Linear Second Order ODE: $y'' + 2 y' + y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 e^{-x} + C_2 x e^{-x}$
+It remains to find a [[Definition:Particular Solution to Differential Equation|particular solution]] $y_p$ to $(1)$.
+Expressing $y_g$ in the form:
+:$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$
+we have:
+{{begin-eqn}}
+{{eqn | l = \map {y_1} x
+      | r = e^{-x}
+      | c = 
+}}
+{{eqn | l = \map {y_2} x
+      | r = x e^{-x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map { {y_1}'} x
+      | r = -e^{-x}
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | l = \map { {y_2}'} x
+      | r = -x e^{-x} + e^{-x}
+      | c = [[Derivative of Exponential Function]], [[Product Rule for Derivatives]]
+}}
+{{end-eqn}}
+By the [[Method of Variation of Parameters]], we have that:
+:$y_p = v_1 y_1 + v_2 y_2$
+where:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+where $\map W {y_1, y_2}$ is the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = {{Defof|Wronskian}}
+}}
+{{eqn | r = e^{-x} \paren {-x e^{-x} + e^{-x} } - x e^{-x} \paren {-e^{-x} }
+      | c = 
+}}
+{{eqn | r = -x e^{-2 x} + e^{-2 x} + x e^{-2 x}
+      | c = 
+}}
+{{eqn | r = e^{-2 x}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int -\frac {x e^{-x} e^{-x} \ln x} {e^{-2 x} } \rd x
+      | c = 
+}}
+{{eqn | r = -\int x \ln x \rd x
+      | c = 
+}}
+{{eqn | r = -\frac {x^2} 2 \paren {\ln x - \frac 1 2}
+      | c = [[Primitive of x by Logarithm of x|Primitive of $x \ln x$]]
+}}
+{{eqn | r = \frac {x^2} 4 - \frac {x^2 \ln x} 2
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {e^{-x} e^{-x} \ln x} {e^{-2 x} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \ln x \rd x
+      | c = 
+}}
+{{eqn | r = x \ln x - x
+      | c = [[Primitive of Logarithm of x|Primitive of $\ln x$]]
+}}
+{{end-eqn}}
+It follows that:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \paren {\frac {x^2} 4 - \frac {x^2 \ln x} 2} e^{-x} + \paren {x \ln x - x} x e^{-x}
+      | c = 
+}}
+{{eqn | r = \frac {x^2 e^{-x} } 4 - \frac {x^2 e^{-x} \ln x} 2 + x^2 e^{-x} \ln x - x^2 e^{-x}
+      | c = simplifying
+}}
+{{eqn | r = \frac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}
+      | c = simplifying
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 e^{-x} + C_2 x e^{-x} - \dfrac {x^2 e^{-x} \ln x} 2 - \dfrac 3 4 x^2 e^{-x}$
+is the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 2 y' - 3 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 2 y' - 3 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{3 x} + C_2 e^{-x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 - 2 m - 3 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = 3$
+:$m_2 = -1$
+These are [[Definition:Real Number|real]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = C_1 e^{3 x} + C_2 e^{-x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+i6015diugly2x7vd35jz1fws8und39v
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' - 2 y' - 3 y = 64 x exp -x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Variation of Parameters
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' - 2 y' - 3 y = 64 x e^{-x}$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 e^{3 x} + C_2 e^{-x} - e^{-x} \paren {8 x^2 + 4 x + 1}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = -2$
+:$q = -3$
+:$\map R x = 64 x e^{-x}$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' - 2 y' - 3 y = 0$
+From [[Linear Second Order ODE/y'' - 2 y' - 3 y = 0|Linear Second Order ODE: $y'' - 2 y' - 3 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 e^{3 x} + C_2 e^{-x}$
+It remains to find a [[Definition:Particular Solution|particular solution]] $y_p$ to $(1)$.
+Expressing $y_g$ in the form:
+:$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$
+we have:
+{{begin-eqn}}
+{{eqn | l = \map {y_1} x
+      | r = e^{3 x}
+      | c = 
+}}
+{{eqn | l = \map {y_2} x
+      | r = e^{-x}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map { {y_1}'} x
+      | r = 3 e^{3 x}
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | l = \map { {y_2}'} x
+      | r = -e^{-x}
+      | c = [[Derivative of Exponential Function]]
+}}
+{{end-eqn}}
+By the [[Method of Variation of Parameters]], we have that:
+:$y_p = v_1 y_1 + v_2 y_2$
+where:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+where $\map W {y_1, y_2}$ is the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = {{Defof|Wronskian}}
+}}
+{{eqn | r = e^{3 x} \paren {-e^{-x} } - e^{-x} \paren {3 e^{3 x} }
+      | c = 
+}}
+{{eqn | r = -e^{2 x} - 3 e^{2 x}
+      | c = 
+}}
+{{eqn | r = -4 e^{2 x}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int -\frac {e^{-x} 64 x e^{-x} } {-4 e^{2 x} } \rd x
+      | c = 
+}}
+{{eqn | r = 16 \int x e^{-4 x} \rd x
+      | c = 
+}}
+{{eqn | r = 16 \frac {e^{-4 x} } {-4} \paren {x - \frac 1 {-4} }
+      | c = [[Primitive of x by Exponential of a x|Primitive of $x e^{a x}$]]
+}}
+{{eqn | r = -4 e^{-4 x} \paren {x + \frac 1 4}
+      | c = 
+}}
+{{eqn | r = -4 x e^{-4 x} - e^{-4 x}
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {e^{3 x} 64 x e^{-x} } {-4 e^{2 x} } \rd x
+      | c = 
+}}
+{{eqn | r = - 16 \int x \rd x
+      | c = 
+}}
+{{eqn | r = -8 x^2
+      | c = [[Primitive of Power]]
+}}
+{{end-eqn}}
+It follows that:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \paren {-4 x e^{-4 x} - e^{-4 x} } e^{3 x} + \paren {-8 x^2} e^{-x}
+      | c = 
+}}
+{{eqn | r = -4 x e^{-x} - e^{-x} - 8 x^2 e^{-x}
+      | c = simplifying
+}}
+{{eqn | r = -e^{-x} \paren {8 x^2 + 4 x + 1}
+      | c = simplifying
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-x} - e^{-x} \paren {8 x^2 + 4 x + 1}$
+is the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 2 y' + 5 y = 0}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 2 y' + 5 y = 0$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad m^2 + 2 m + 5 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = -1 + 2 i$
+:$m_2 = -1 - 2 i$
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+tn73yr2w61wblyer24fduyyaxuq2kvc
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 2 y' + 5 y = exp -x secant 2 x}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Variation of Parameters
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 2 y' + 5 y = e^{-x} \sec 2 x$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \dfrac {x e^{-x} \sin 2 x} 2 + \dfrac {e^{-x} \cos 2 x \ln \cos 2 x} 4$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 2$
+:$q = 5$
+:$\map R x = e^{-x} \sec 2 x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' + 2 y' + 5 y = 0$
+From [[Linear Second Order ODE/y'' + 2 y' + 5 y = 0|Linear Second Order ODE: $y'' + 2 y' + 5 y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x}$
+It remains to find a [[Definition:Particular Solution to Differential Equation|particular solution]] $y_p$ to $(1)$.
+Expressing $y_g$ in the form:
+:$y_g = C_1 \, \map {y_1} x + C_2 \, \map {y_2} x$
+we have:
+{{begin-eqn}}
+{{eqn | l = \map {y_1} x
+      | r = e^{-x} \cos 2 x
+      | c = 
+}}
+{{eqn | l = \map {y_2} x
+      | r = e^{-x} \sin 2 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map { {y_1}'} x
+      | r = -e^{-x} \cos 2 x - 2 e^{-x} \sin 2 x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | r = -e^{-x} \paren {\cos 2 x + 2 \sin 2 x}
+      | c = 
+}}
+{{eqn | l = \map { {y_2}'} x
+      | r = -e^{-x} \sin 2 x + 2 e^{-x} \cos 2 x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | r = e^{-x} \paren {2 \cos 2 x - \sin 2 x}
+      | c = 
+}}
+{{end-eqn}}
+By the [[Method of Variation of Parameters]], we have that:
+:$y_p = v_1 y_1 + v_2 y_2$
+where:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+where $\map W {y_1, y_2}$ is the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = {{Defof|Wronskian}}
+}}
+{{eqn | r = e^{-x} \cos 2 x \paren {e^{-x} \paren {2 \cos 2 x - \sin 2 x} } - e^{-x} \sin 2 x \paren {-e^{-x} \paren {\cos 2 x + 2 \sin 2 x} }
+      | c = 
+}}
+{{eqn | r = e^{-2 x} \paren {2 \cos^2 2 x - \cos 2 x \sin 2 x + \sin 2 x \cos 2 x + 2 \sin^2 2 x}
+      | c = 
+}}
+{{eqn | r = 2 e^{-2 x} \paren {\cos^2 2 x + \sin^2 2 x}
+      | c = 
+}}
+{{eqn | r = 2 e^{-2 x}
+      | c = [[Sum of Squares of Sine and Cosine]]
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int -\frac {e^{-x} \sin 2 x e^{-x} \sec 2 x} {2 e^{-2 x} } \rd x
+      | c = 
+}}
+{{eqn | r = -\frac 1 2 \int \sin 2 x \sec 2 x \rd x
+      | c = 
+}}
+{{eqn | r = -\frac 1 2 \int \frac {\sin 2 x} {\cos 2 x} \rd x
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = -\frac 1 2 \int \tan 2 x \rd x
+      | c = [[Tangent is Sine divided by Cosine]]
+}}
+{{eqn | r = \frac 1 4 \ln \cos 2 x
+      | c = [[Primitive of Tangent of a x|Primitive of $\tan a x$]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \, \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {e^{-x} \cos 2 x e^{-x} \sec 2 x} {2 e^{-2 x} } \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \int \cos 2 x \sec 2 x \rd x
+      | c = 
+}}
+{{eqn | r = \frac 1 2 \int \rd x
+      | c = [[Secant is Reciprocal of Cosine]]
+}}
+{{eqn | r = \frac x 2
+      | c = [[Primitive of Constant]]
+}}
+{{end-eqn}}
+It follows that:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \paren {\frac 1 4 \ln \cos 2 x} e^{-x} \cos 2 x + \paren {\frac x 2} e^{-x} \sin 2 x
+      | c = 
+}}
+{{eqn | r = \frac {x e^{-x} \sin 2 x} 2 + \frac {e^{-x} \cos 2 x \ln \cos 2 x} 4
+      | c = rearranging
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = e^{-x} \paren {C_1 \cos 2 x + C_2 \sin 2 x} + \dfrac {x e^{-x} \sin 2 x} 2 + \dfrac {e^{-x} \cos 2 x \ln \cos 2 x} 4$
+is the [[Definition:General Solution to Differential Equation|general solution]] to $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Singleton is Directed and Filtered Subset}
+Tags: Preorder Theory
+
+\begin{theorem}
+Let $\struct {S, \precsim}$ be a [[Definition:Preordered Set|preordered set]].
+Let $x$ be an [[Definition:Element|element]] of $S$.
+Then $\set x$ is [[Definition:Directed Subset|directed]] and [[Definition:Filtered Subset|filtered subset]] of $S$.
+\end{theorem}
+
+\begin{proof}
+Let $y, z \in \set x$.
+By definition of [[Definition:Singleton|singleton]]:
+:$ y = x \land z = x$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$y \precsim x \land z \precsim x$
+Thus:
+:$\exists h \in \set x: y \precsim h \land z \precsim h$
+Thus by definition:
+:$\set x$ is a [[Definition:Directed Subset|directed subset]] of $S$.
+$\set x$ is a [[Definition:Filtered Subset|filtered subset]] of $S$ by [[Definition:Mutatis Mutandis|mutatis mutandis]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Second Order ODE/(x^2 - 1) y'' - 2 x y' + 2 y = 0}
+Tags: Examples of Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad \paren {x^2 - 1} y'' - 2 x y' + 2 y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 x + C_2 \paren {x^2 + 1}$
+\end{theorem}
+
+\begin{proof}
+Note that:
+{{begin-eqn}}
+{{eqn | l = y_1
+      | r = x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}'
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_1}''
+      | r = 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+and so by inspection:
+:$y_1 = x$
+is a [[Definition:Particular Solution|particular solution]] of $(1)$.
+$(1)$ can be expressed as:
+:$(2): \quad y'' - \dfrac {2 x} {x^2 - 1} y' + \dfrac 2 {x^2 - 1} y = 0$
+which is in the form:
+:$y'' + \map P x y' + \map Q x y = 0$
+where:
+:$\map P x = - \dfrac {2 x} {x^2 - 1}$
+:$\map Q x = \dfrac 2 {x^2 - 1}$
+From [[Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another]]:
+:$\map {y_2} x = \map v x \, \map {y_1} x$
+where:
+:$\displaystyle v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$
+is also a [[Definition:Particular Solution|particular solution]] of $(1)$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \int P \rd x
+      | r = \int \paren {- \dfrac {2 x} {x^2 - 1} } \rd x
+      | c = 
+}}
+{{eqn | r = -\map \ln {x^2 - 1}
+      | c = [[Primitive of Function under its Derivative]]
+}}
+{{eqn | ll= \leadsto
+      | l = e^{-\int P \rd x}
+      | r = e^{\map \ln {x^2 - 1} }
+      | c = 
+}}
+{{eqn | r = x^2 - 1
+      | c =  
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v
+      | r = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x
+      | c = Definition of $v$
+}}
+{{eqn | r = \int \dfrac 1 {x^2} \paren {x^2 - 1} \rd x
+      | c = 
+}}
+{{eqn | r = \int \paren {1 - \dfrac 1 {x^2} } \rd x
+      | c = 
+}}
+{{eqn | r = x + \frac 1 x
+      | c = [[Primitive of Power]]
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_2
+      | r = v y_1
+      | c = Definition of $y_2$
+}}
+{{eqn | r = \paren {x + \frac 1 x} x
+      | c = 
+}}
+{{eqn | r = x^2 + 1
+      | c = 
+}}
+{{end-eqn}}
+From [[Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution]]:
+:$y = C_1 x + C_2 \paren {x^2 + 1}$
+{{qed}}
+[[Category:Examples of Homogeneous LSOODEs]]
+qg4jptdxydcikqwktj9ed8dkcpb9j3n
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/(x^2 - 1) y'' - 2 x y' + 2 y = (x^2 - 1)^2}
+Tags: Examples of Linear Second Order ODEs, Examples of Method of Variation of Parameters
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad \paren {x^2 - 1} y'' - 2 x y' + 2 y = \paren {x^2 - 1}^2$
+has the [[Definition:General Solution|general solution]]:
+:$y = C_1 x + C_2 \paren {x^2 + 1} + \dfrac {x^4} 6 - \dfrac {x^2} 2$
+\end{theorem}
+
+\begin{proof}
+$(1)$ can be manipulated into the form:
+:$y'' - \dfrac {2 x} {x^2 - 1} y' + \dfrac 2 {x^2 - 1} y = x^2 - 1$
+It can be seen that this is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] in the form:
+:$y'' + \map P x y' + \map Q x y = \map R x$
+where:
+:$\map P x = -\dfrac {2 x} {x^2 - 1}$
+:$\map Q x = \dfrac 2 {x^2 - 1}$
+:$\map R x = x^2 - 1$
+First we establish the solution of the corresponding [[Definition:Homogeneous Linear Second Order ODE|homogeneous linear second order ODE]]:
+:$\paren {x^2 - 1} y'' - 2 x y' + 2 y = 0$
+From [[Second Order ODE/(x^2 - 1) y'' - 2 x y' + 2 y = 0|Second Order ODE: $\paren {x^2 - 1} y'' - 2 x y' + 2 y = 0$]], this has the [[Definition:General Solution|general solution]]:
+:$y_g = C_1 x + C_2 \paren {x^2 + 1}$
+It remains to find a [[Definition:Particular Solution|particular solution]] $y_p$ to $(1)$.
+Expressing $y_g$ in the form:
+:$y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$
+we have:
+{{begin-eqn}}
+{{eqn | l = \map {y_1} x
+      | r = x
+      | c = 
+}}
+{{eqn | l = \map {y_2} x
+      | r = x^2 + 1
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map { {y_1}'} x
+      | r = 1
+      | c = [[Power Rule for Derivatives]]
+}}
+{{eqn | l = \map { {y_2}'} x
+      | r = 2 x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+By the [[Method of Variation of Parameters]], we have that:
+:$y_p = v_1 y_1 + v_2 y_2$
+where:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+where $\map W {y_1, y_2}$ is the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = {{Defof|Wronskian}}
+}}
+{{eqn | r = x \paren {2 x} - \paren {x^2 + 1} \paren 1
+      | c = 
+}}
+{{eqn | r = 2 x^2 - x^2 - 1
+      | c = 
+}}
+{{eqn | r = x^2 - 1
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int -\frac {\paren {x^2 + 1} \paren {x^2 - 1} } {x^2 - 1} \rd x
+      | c = 
+}}
+{{eqn | r = -\int \paren {x^2 + 1} \rd x
+      | c = 
+}}
+{{eqn | r = -\frac {x^3} 3 - x
+      | c = [[Primitive of Power]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {x \paren {x^2 - 1} } {x^2 - 1} \rd x
+      | c = 
+}}
+{{eqn | r = \int x \rd x
+      | c = 
+}}
+{{eqn | r = \frac {x^2} 2
+      | c = [[Primitive of Power]]
+}}
+{{end-eqn}}
+It follows that:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \paren {-\frac {x^3} 3 - x} x + \paren {\frac {x^2} 2} \paren {x^2 + 1}
+      | c = 
+}}
+{{eqn | r = -\frac {x^4} 3 - x^2 + \frac {x^4} 2 + \frac {x^2} 2
+      | c = rearranging
+}}
+{{eqn | r = \frac {x^4} 6 - \frac {x^2} 2
+      | c = rearranging
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 x + C_2 \paren {x^2 + 1} + \dfrac {x^4} 6 - \dfrac {x^2} 2$
+is the [[Definition:General Solution|general solution]] to $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Directed iff Lower Closure Directed}
+Tags: Preorder Theory
+
+\begin{theorem}
+Let $\left({S, \precsim}\right)$ be a [[Definition:Preordered Set|preordered set]].
+Let $H$ be a [[Definition:Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Then $H$ is [[Definition:Directed Subset|directed]] {{iff}}
+$H^\precsim$ is [[Definition:Directed Subset|directed]]
+where $H^\precsim$ denotes the [[Definition:Lower Closure/Set|lower closure of set]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let us assume that
+:$H$ is [[Definition:Directed Subset|directed]].
+Let $x, y \in H^\precsim$
+By definition of [[Definition:Lower Closure/Set|lower closure]]:
+:$\exists x' \in H: x \precsim x'$
+and
+:$\exists y' \in H: y \precsim y'$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists z \in H: x' \precsim z \land y' \precsim z$
+By definition of [[Definition:Reflexivity|reflexivity]];
+:$z \precsim z$
+By definition of [[Definition:Lower Closure/Set|lower closure]]:
+:$z \in H^\precsim$
+Thus by definition of [[Definition:Transitivity|transitivity]]
+:$\exists z \in H^\precsim: x \precsim z \land y \precsim z$
+Thus by definition
+:$H^\precsim$ is [[Definition:Directed Subset|directed]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Let us assume that
+:$H^\precsim$ is [[Definition:Directed Subset|directed]].
+Let $x, y \in H$.
+By definition of [[Definition:Reflexivity|reflexivity]];
+:$x \precsim x \land y \precsim y$
+By definition of [[Definition:Lower Closure/Set|lower closure]]:
+:$x, y \in H^\precsim$
+By definition of [[Definition:Directed Subset|directed subset]]:
+:$\exists z \in H^\precsim: x \precsim z \land y \precsim z$
+By definition of [[Definition:Lower Closure/Set|lower closure]]:
+:$\exists z' \in H: z \precsim z'$
+Thus by definition of [[Definition:Transitivity|transitivity]]
+:$\exists z' \in H: x \precsim z' \land y \precsim z'$
+Thus by definition
+:$H$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Filtered iff Upper Closure Filtered}
+Tags: Preorder Theory
+
+\begin{theorem}
+Let $\left({S, \precsim}\right)$ be a [[Definition:Preordered Set|preordered set]].
+Let $H$ be a [[Definition:Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
+Then $H$ is [[Definition:Filtered Subset|filtered]] {{iff}}
+$H^\succsim$ is [[Definition:Filtered Subset|filtered]]
+where $H^\succsim$ denotes the [[Definition:Upper Closure/Set|upper closure of set]].
+\end{theorem}
+
+\begin{proof}
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]] of the proof of [[Directed iff Lower Closure Directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0}
+Tags: Examples of Homogeneous LSOODEs, Linear Second Order ODE: (x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad \paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$
+has the [[Definition:General Solution of Differential Equation|general solution]]:
+:$y = C_1 e^x + \dfrac {C_2} x$
+\end{theorem}<|endoftext|>
+\section{Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = x (x + 1)^2}
+Tags: Examples of Linear Second Order ODEs, Examples of Method of Variation of Parameters
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad \paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = x \paren {x + 1}^2$
+has the [[Definition:General Solution|general solution]]:
+:$y = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$
+\end{theorem}
+
+\begin{proof}
+$(1)$ can be manipulated into the form:
+:$y'' + \dfrac {2 - x^2} {x^2 + x} y' - \dfrac {2 + x} {x^2 + x} y = x + 1$
+It can be seen that this is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] in the form:
+:$y'' + \map P x y' + \map Q x y = \map R x$
+where:
+:$\map P x = \dfrac {2 - x^2} {x^2 + x}$
+:$\map Q x = -\dfrac {2 + x} {x^2 + x}$
+:$\map R x = x + 1$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$\paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$
+From [[Linear Second Order ODE/(x^2 + x) y'' + (2 - x^2) y' - (2 + x) y = 0|Linear Second Order ODE: $\paren {x^2 + x} y'' + \paren {2 - x^2} y' - \paren {2 + x} y = 0$]], this has the [[Definition:General Solution|general solution]]:
+:$y_g = C_1 e^x + \dfrac {C_2} x$
+It remains to find a [[Definition:Particular Solution|particular solution]] $y_p$ to $(1)$.
+Expressing $y_g$ in the form:
+:$y_g = C_1 \map {y_1} x + C_2 \map {y_2} x$
+we have:
+{{begin-eqn}}
+{{eqn | l = \map {y_1} x
+      | r = e^x
+      | c = 
+}}
+{{eqn | l =\map {y_2} x
+      | r = \dfrac 1 x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \map { {y_1}'} x
+      | r = e^x
+      | c = [[Derivative of Exponential Function]]
+}}
+{{eqn | l = \map { {y_2}'} x
+      | r = -\dfrac 1 {x^2}
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+By the [[Method of Variation of Parameters]], we have that:
+:$y_p = v_1 y_1 + v_2 y_2$
+where:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{end-eqn}}
+where $\map W {y_1, y_2}$ is the [[Definition:Wronskian|Wronskian]] of $y_1$ and $y_2$.
+We have that:
+{{begin-eqn}}
+{{eqn | l = \map W {y_1, y_2}
+      | r = y_1 {y_2}' - y_2 {y_1}'
+      | c = {{Defof|Wronskian}}
+}}
+{{eqn | r = e^x \paren {-\dfrac 1 {x^2} } - \paren {\dfrac 1 x} \paren {e^x}
+      | c = 
+}}
+{{eqn | r = -\dfrac {e^x} {x^2} - \dfrac {e^x} x
+      | c = 
+}}
+{{eqn | r = -e^x \, \dfrac {x + 1} {x^2}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = v_1
+      | r = \int -\frac {y_2 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int -\frac {\paren {\dfrac 1 x} \paren {x + 1} } {-e^x \, \dfrac {x + 1} {x^2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int x e^{-x} \rd x
+      | c = 
+}}
+{{eqn | r = -e^{-x} \paren {x + 1}
+      | c = [[Primitive of x by Exponential of a x|Primitive of $x e^{a x}$]]
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = v_2
+      | r = \int \frac {y_1 \map R x} {\map W {y_1, y_2} } \rd x
+      | c = 
+}}
+{{eqn | r = \int \frac {e^x \paren {x + 1} } {-e^x \, \dfrac {x + 1} {x^2} } \rd x
+      | c = 
+}}
+{{eqn | r = -\int x^2 \rd x
+      | c = 
+}}
+{{eqn | r = -\frac {x^3} 3
+      | c = [[Primitive of Power]]
+}}
+{{end-eqn}}
+It follows that:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \paren {-e^{-x} \paren {x + 1} } e^x + \paren {-\frac {x^3} 3} \paren {\dfrac 1 x}
+      | c = 
+}}
+{{eqn | r = -x - 1 - \frac {x^2} 3
+      | c = rearranging
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 e^x + \dfrac {C_2} x - x - 1 - \dfrac {x^2} 3$
+is the [[Definition:General Solution|general solution]] to $(1)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Position of Cart attached to Wall by Spring/x = x0 at t = 0}
+Tags: Mechanics
+
+\begin{theorem}
+Let $C$ be pulled aside to $x = x_0$ and released from [[Definition:Stationary|stationary]] at time $t = 0$.
+Then the [[Definition:Horizontal|horizontal]] [[Definition:Position|position]] of $C$ at [[Definition:Time|time]] $t$ can be expressed as:
+:$x = x_0 \cos \alpha t$
+\end{theorem}
+
+\begin{proof}
+From [[Position of Cart attached to Wall by Spring]], the [[Definition:Horizontal|horizontal]] [[Definition:Position|position]] of $C$ is given as:
+:$(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$
+where:
+:$C_1$ and $C_2$ depend upon the [[Definition:Initial Condition|conditions]] of $C$ at [[Definition:Time|time]] $t = 0$
+:$\alpha = \sqrt {\dfrac k m}$
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT}} $x$ gives:
+:$(2): \quad x' = -C_1 \alpha \sin \alpha t + C_2 \alpha \cos \alpha t$
+Setting the [[Definition:Initial Condition|initial condition]] $x = x_0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = x_0
+      | r = C_1 \cos 0 + C_2 \sin 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = x_0
+      | r = C_1
+      | c = 
+}}
+{{end-eqn}}
+Setting the [[Definition:Initial Condition|initial condition]] $x' = 0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = -C_1 \alpha \sin 0 + C_2 \alpha \cos 0
+      | c = 
+}}
+{{eqn | ll= \implies
+      | l = 0
+      | r = C_2
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+:$x = x_0 \cos \alpha t$
+where $\alpha = \sqrt {\dfrac k m}$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Directed in Join Semilattice}
+Tags: Join and Meet Semilattices, Lower Sets
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
+Let $H$ be a [[Definition:Empty Set|non-empty]] [[Definition:Lower Set|lower]] [[Definition:Subset|subset]] of $S$.
+Then $H$ is [[Definition:Directed Subset|directed]] {{iff}}
+:$\forall x, y \in H: x \vee y \in H$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let us assume that
+:$H$ is [[Definition:Directed Subset|directed]].
+Let $x, y \in H$.
+By definition of [[Definition:Directed Subset|directed]]:
+:$\exists z \in H: x \preceq z \land y \preceq z$
+By definition
+:$z$ is [[Definition:Upper Bound of Set|upper bound]] of $\left\{ {x, y}\right\}$
+By definitions of [[Definition:Supremum of Set|supremum]] and [[Definition:Join (Order Theory)|join]]:
+:$x \vee y = \sup\left\{ {x, y}\right\} \preceq z$
+Thus by definition of [[Definition:Lower Set|lower set]]:
+:$x \vee y \in H$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let us assume that
+:$\forall x, y \in H: x \vee y \in H$
+Let $x, y \in H$.
+By assumption:
+:$x \vee y \in H$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$x \vee y$ is [[Definition:Upper Bound of Set|upper bound]] of $\left\{ {x, y}\right\}$
+Thus
+:$x \preceq x \vee y \land y \preceq x \vee y$
+Thus by definition
+:$H$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Position of Cart attached to Wall by Spring under Damping/Overdamped/x = x0 at t = 0}
+Tags: Mechanics
+
+\begin{theorem}
+Let $C$ be pulled aside to $x = x_0$ and released from [[Definition:Stationary|stationary]] at time $t = 0$.
+Then the [[Definition:Horizontal|horizontal]] [[Definition:Position|position]] of $C$ at [[Definition:Time|time]] $t$ can be expressed as:
+:$x = \dfrac {x_0} {m_1 - m_2} \left({m_1 e^{m_2 t} - m_2 e^{m_1 t} }\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Position of Cart attached to Wall by Spring under Damping/Overdamped|Position of Cart attached to Wall by Spring under Damping: Overdamped]]:
+:$(1): \quad x = C_1 e^{m_1 t} + C_2 e^{m_2 t}$
+where $C_1$ and $C_2$ depend upon the [[Definition:Initial Condition|conditions]] of $C$ at [[Definition:Time|time]] $t = 0$.
+It remains to evaluate $C_1$ and $C_2$ under the given [[Definition:Initial Condition|conditions]].
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT}} $t$ gives:
+:$(2): \quad x' = C_1 m_1 e^{m_1 t} + C_2 m_2 e^{m_2 t}$
+Setting the [[Definition:Initial Condition|initial condition]] $x = x_0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = x_0
+      | r = C_1 e^0 + C_2 e^0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_0
+      | r = C_1 + C_2
+      | c = 
+}}
+{{end-eqn}}
+Setting the [[Definition:Initial Condition|initial condition]] $x' = 0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = C_1 m_1 e^0 + C_2 m_2 e^0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | r = C_1 m_1 + C_2 m_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | r = C_1 m_1 + \left({x_0 - C_1}\right) m_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_1 \left({m_1 - m_2}\right)
+      | r = -x_0 m_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_1 \left({m_1 - m_2}\right)
+      | r = -\dfrac {x_0 m_2} {m_1 - m_2}
+      | c = 
+}}
+{{end-eqn}}
+Then we have:
+{{begin-eqn}}
+{{eqn | l = x_0
+      | r = C_1 + C_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_2
+      | r = x_0 - C_1
+      | c = 
+}}
+{{eqn | r = x_0 + \dfrac {x_0 m_2} {m_1 - m_2}
+      | c = 
+}}
+{{eqn | r = x_0 \left({1 + \dfrac {m_2} {m_1 - m_2} }\right)
+      | c = 
+}}
+{{eqn | r = x_0 \left({\dfrac {\left({m_1 - m_2}\right) + m_2} {m_1 - m_2} }\right)
+      | c = 
+}}
+{{eqn | r = \dfrac {x_0 m_1} {m_1 - m_2}
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = -\dfrac {x_0 m_2} {m_1 - m_2} e^{m_1 t} + \dfrac {x_0 m_1} {m_1 - m_2} e^{m_2 t}
+      | c = 
+}}
+{{eqn | r = \dfrac {x_0} {m_1 - m_2} \left({m_1 e^{m_2 t} - m_2 e^{m_1 t} }\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Position of Cart attached to Wall by Spring under Damping/Critically Damped/x = x0 at t = 0}
+Tags: Mechanics
+
+\begin{theorem}
+Let $C$ be pulled aside to $x = x_0$ and released from [[Definition:Stationary|stationary]] at time $t = 0$.
+Then the [[Definition:Horizontal|horizontal]] [[Definition:Position|position]] of $C$ at [[Definition:Time|time]] $t$ can be expressed as:
+:$x = x_0 e^{-a t} \left({1 + a t}\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Position of Cart attached to Wall by Spring under Damping/Critically Damped|Position of Cart attached to Wall by Spring under Damping: Critically Damped]]:
+:$x = C_1 e^{-a t} + C_2 t e^{-a t}$
+where $C_1$ and $C_2$ depend upon the [[Definition:Initial Condition|conditions]] of $C$ at [[Definition:Time|time]] $t = 0$.
+It remains to evaluate $C_1$ and $C_2$ under the given [[Definition:Initial Condition|conditions]].
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT}} $t$ gives:
+:$(2): \quad x' = -a C_1 e^{-a t} + C_2 e^{-a t} - a C_2 t e^{-a t}$
+Setting the [[Definition:Initial Condition|initial condition]] $x = x_0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = x_0
+      | r = C_1 e^0 + C_2 e^0 \times 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = x_0
+      | r = C_1
+      | c = 
+}}
+{{end-eqn}}
+Setting the [[Definition:Initial Condition|initial condition]] $x' = 0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = -a C_1 e^0 + C_2 e^0 - a C_2 e^0 \times 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = 0
+      | r = -a C_1 + C_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_2
+      | r = a C_1 
+      | c = 
+}}
+{{eqn | r = a x_0
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = x_0 e^{-a t} + a x_0 t e^{-a t}
+      | c = 
+}}
+{{eqn | r = x_0 e^{-a t} \left({1 + a t}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Position of Cart attached to Wall by Spring under Damping/Underdamped/x = x0 at t = 0}
+Tags: Mechanics
+
+\begin{theorem}
+Let $C$ be pulled aside to $x = x_0$ and released from [[Definition:Stationary|stationary]] at time $t = 0$.
+Then the [[Definition:Horizontal|horizontal]] [[Definition:Position|position]] of $C$ at [[Definition:Time|time]] $t$ can be expressed as:
+:$x = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Position of Cart attached to Wall by Spring under Damping/Underdamped|Position of Cart attached to Wall by Spring under Damping: Underdamped]]:
+:$(2): \quad x = e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right)$
+where $C_1$ and $C_2$ depend upon the [[Definition:Initial Condition|conditions]] of $C$ at [[Definition:Time|time]] $t = 0$.
+It remains to evaluate $C_1$ and $C_2$ under the given [[Definition:Initial Condition|conditions]].
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT}} $t$ gives:
+:$(2): \quad x' = -b e^{-b t} \left({C_1 \cos \alpha t + C_2 \sin \alpha t}\right) + e^{-b t} \left({-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}\right)$
+Setting the [[Definition:Initial Condition|initial condition]] $x = x_0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = x_0
+      | r = e^0 \left({C_1 \cos 0 + C_2 \sin 0}\right)
+      | c = 
+}}
+{{eqn | r = C_1
+      | c = 
+}}
+{{end-eqn}}
+Setting the [[Definition:Initial Condition|initial condition]] $x' = 0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = -b e^0 \left({C_1 \cos 0 + C_2 \sin 0}\right) + e^0 \left({-\alpha C_1 \sin 0 + \alpha C_2 \cos 0}\right)
+      | c = 
+}}
+{{eqn | r = -b C_1 + \alpha C_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_2
+      | r = \frac {b C_1} \alpha
+      | c = 
+}}
+{{eqn | r = \frac {b x_0} \alpha
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = e^{-b t} \left({x_0 \cos \alpha t + \frac {b x_0} \alpha \sin \alpha t}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Canonical Form of Underdamped Oscillatory System}
+Tags: Mechanics
+
+\begin{theorem}
+Consider a [[Definition:Physical System|physical system]] $S$ whose behaviour can be described with the [[Definition:Second Order ODE|second order ODE]] in the form:
+:$(1): \quad \dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
+for $a, b \in \R_{>0}$.
+Let $b < a$, so as to make $S$ [[Definition:Underdamped|underdamped]].
+Then the value of $x$ can be expressed in the form:
+:$x = \dfrac {x_0 \, a} \alpha e^{-b t} \, \map \cos {\alpha t - \theta}$
+where:
+:$\alpha = \sqrt {a^2 - b^2}$
+:$\theta = \map \arctan {\dfrac b \alpha}$
+This can be referred to as the '''canonical form''' of the [[Definition:General Solution|solution]] of $(1)$.
+\end{theorem}
+
+\begin{proof}
+From [[Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation|Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation]], the [[Definition:General Solution|general solution]] of $(1)$ is:
+:$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$
+where:
+:$\alpha = \sqrt {a^2 - b^2}$
+This is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|homogeneous linear second order ODE with constant coefficients]].
+Let $m_1$ and $m_2$ be the [[Definition:Root of Polynomial|roots]] of the [[Definition:Auxiliary Equation|auxiliary equation]]:
+:$m^2 + 2 b + a^2 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]]:
+{{begin-eqn}}
+{{eqn | l = m_1
+      | r = -b + i \sqrt {a^2 - b^2}
+      | c = 
+}}
+{{eqn | l = m_1, m_2
+      | r = -b - i \sqrt {a^2 - b^2}
+      | c = 
+}}
+{{end-eqn}}
+So from [[Solution of Constant Coefficient Homogeneous LSOODE/Complex Roots of Auxiliary Equation|Solution of Constant Coefficient Homogeneous LSOODE: Complex Roots of Auxiliary Equation]]:
+:$x = e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t}$
+where:
+:$\alpha = \sqrt {a^2 - b^2}$
+The following assumptions are made:
+:We may label a particular point in [[Definition:Time|time]] $t = 0$ at which the [[Definition:Derivative|derivative]] of $x$ {{WRT}} $t$ is itself [[Definition:Zero|zero]].
+:We allow that at this arbitrary $t = 0$, the value of $x$ is assigned the value $x = x_0$.
+This corresponds, for example, with a [[Definition:Physical System|physical system]] in which the moving [[Definition:Body|body]] is pulled from its [[Definition:Equilibrium Position of Spring|equilibrium position]] and released from [[Definition:Stationary|stationary]] at [[Definition:Time|time]] [[Definition:Zero|zero]].
+[[Definition:Differentiation|Differentiating]] $(1)$ {{WRT}} $t$ gives:
+:$\quad x' = -b e^{-b t} \paren {C_1 \cos \alpha t + C_2 \sin \alpha t} + e^{-b t} \paren {-\alpha C_1 \sin \alpha t + \alpha C_2 \cos \alpha t}$
+Setting the [[Definition:Initial Condition|initial condition]] $x = x_0$ when $t = 0$ in $(1)$:
+{{begin-eqn}}
+{{eqn | l = x_0
+      | r = e^0 \paren {C_1 \cos 0 + C_2 \sin 0}
+      | c = 
+}}
+{{eqn | r = C_1
+      | c = 
+}}
+{{end-eqn}}
+Setting the [[Definition:Initial Condition|initial condition]] $x' = 0$ when $t = 0$:
+{{begin-eqn}}
+{{eqn | l = 0
+      | r = -b e^0 \paren {C_1 \cos 0 + C_2 \sin 0} + e^0 \paren {-\alpha C_1 \sin 0 + \alpha C_2 \cos 0}
+      | c = 
+}}
+{{eqn | r = -b C_1 + \alpha C_2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = C_2
+      | r = \frac {b C_1} \alpha
+      | c = 
+}}
+{{eqn | r = \frac {b x_0} \alpha
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = e^{-b t} \paren {x_0 \cos \alpha t + \frac {b x_0} \alpha \sin \alpha t}
+      | c = 
+}}
+{{eqn | n = 2
+      | r = e^{-b t} \dfrac {x_0} \alpha \paren {\alpha \cos \alpha t + b \sin \alpha t}
+      | c = 
+}}
+{{end-eqn}}
+From [[Multiple of Sine plus Multiple of Cosine/Cosine Form|Multiple of Sine plus Multiple of Cosine:Cosine Form]], $(2)$ can be expressed as:
+{{begin-eqn}}
+{{eqn | l = x
+      | r = \dfrac {x_0} \alpha e^{-b t} \paren {\sqrt {\alpha^2 + b^2} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha} }
+      | c = 
+}}
+{{eqn | r = \dfrac {x_0 \sqrt {\sqrt{a^2 - b^2}^2 + b^2} } \alpha e^{-b t} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha}
+      | c = 
+}}
+{{eqn | r = \dfrac {x_0 \sqrt {a^2 - b^2 + b^2} } \alpha e^{-b t} \, \map \cos {\alpha t + \arctan \dfrac {-b} \alpha}
+      | c = 
+}}
+{{eqn | r = \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t + \arctan \dfrac {-b} \alpha}
+      | c = 
+}}
+{{eqn | r = \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t - \arctan \dfrac b \alpha}
+      | c = [[Tangent Function is Odd]]
+}}
+{{eqn | r = \dfrac {x_0 \, a} \alpha e^{-b t} \, \cos {\alpha t - \theta}
+      | c = where $\theta = \arctan \dfrac b \alpha$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Period of Oscillation of Underdamped System is Regular}
+Tags: Mechanics
+
+\begin{theorem}
+Consider a [[Definition:Physical System|physical system]] $S$ whose behaviour can be described with the [[Definition:Second Order ODE|second order ODE]] in the form:
+:$\dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
+for $a, b \in \R_{>0}$.
+Let $b < a$, so as to make $S$ [[Definition:Underdamped|underdamped]].
+Then the [[Definition:Period of Underdamped Oscillation|period]] of its movement is well-defined, in the sense that its [[Definition:Zero of Function|zeroes]] are regularly spaced, and given by:
+:$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$
+\end{theorem}
+
+\begin{proof}
+Let the position of $S$ be described in the [[Canonical Form of Underdamped Oscillatory System|canonical form]]:
+:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
+where $\alpha = \sqrt {a^2 - b^2}$.
+The [[Definition:Zero of Function|zeroes]] of $(1)$ occur exactly where:
+:$\map \cos {\alpha t - \theta} = 0$
+Thus the [[Definition:Period of Function|period]] $T$ of $\map \cos {\alpha t - \theta}$ is given by:
+:$\alpha T = 2 \pi$
+and so:
+:$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Period of Oscillation of Underdamped Cart attached to Wall by Spring}
+Tags: Mechanics
+
+\begin{theorem}
+Let $C$ be pulled aside to $x = x_0$ and released from [[Definition:Stationary|stationary]] at time $t = 0$.
+Then the [[Definition:Period of Underdamped Oscillation|period of oscillation]] of $C$ can be expressed as:
+:$T = \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }$
+\end{theorem}
+
+\begin{proof}
+Let:
+:$a^2 = \dfrac k m$
+:$2 b = \dfrac c m$
+From [[Position of Cart attached to Wall by Spring under Damping/Underdamped/x = x0 at t = 0|Position of Cart attached to Wall by Spring under Damping: Underdamped: $x = x_0$ at $t = 0$]]:
+:$x = \dfrac {x_0} \alpha e^{-b t} \left({\alpha \cos \alpha t + b \sin \alpha t}\right)$
+where $\alpha = \sqrt {a^2 - b^2}$.
+Let $T$ be the [[Definition:Period of Underdamped Oscillation|period of oscillation]] of $C$.
+Then:
+{{begin-eqn}}
+{{eqn | l = T
+      | r = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }
+      | c = [[Period of Oscillation of Underdamped System is Regular]]
+}}
+{{eqn | r = \dfrac {2 \pi} {\sqrt {\dfrac k m - \left({\dfrac c {2 m} }\right)^2} }
+      | c = 
+}}
+{{eqn | r = \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Frequency of Underdamped System}
+Tags: Mechanics
+
+\begin{theorem}
+Consider a [[Definition:Physical System|physical system]] $S$ whose behaviour can be described with the [[Definition:Second Order ODE|second order ODE]] in the form:
+:$\dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
+for $a, b \in \R_{>0}$.
+Let $b < a$, so as to make $S$ [[Definition:Underdamped|underdamped]].
+Then the [[Definition:Natural Frequency|natural frequency]] of $S$ is given by:
+:$\nu = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$
+\end{theorem}
+
+\begin{proof}
+Let the position of $S$ be described in the [[Canonical Form of Underdamped Oscillatory System|canonical form]]:
+:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
+where $\alpha = \sqrt {a^2 - b^2}$.
+From [[Period of Oscillation of Underdamped System is Regular]]:
+:$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$
+where $T$ is the [[Definition:Period of Underdamped Oscillation|period]] of the movement of $S$.
+By definition of [[Definition:Natural Frequency|natural frequency]]:
+:$\nu = \dfrac 1 T = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Natural Frequency of Underdamped Cart attached to Wall by Spring}
+Tags: Mechanics
+
+\begin{theorem}
+Let $C$ be pulled aside to $x = x_0$ and released from [[Definition:Stationary|stationary]] at time $t = 0$.
+Then the [[Definition:Natural Frequency|natural frequency]] of $C$ can be expressed as:
+:$\nu = \dfrac 1 {2 \pi} \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$
+\end{theorem}
+
+\begin{proof}
+From [[Period of Oscillation of Underdamped Cart attached to Wall by Spring]]:
+:$T = \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }$
+where $T$ is the [[Definition:Period of Underdamped Oscillation|period of oscillation]] of $C$.
+Then by definition of the [[Definition:Natural Frequency|natural frequency]] of $C$:
+{{begin-eqn}}
+{{eqn | l = \nu
+      | r = \dfrac 1 T
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {\left({\dfrac {2 \pi} {\sqrt {\frac k m - \frac {c^2} {4 m^2} } } }\right)}
+      | c = 
+}}
+{{eqn | r = \dfrac 1 {2 \pi} \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 2 b y' + a^2 y = 0/b less than a}
+Tags: Examples of Constant Coefficient Homogeneous LSOODEs
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 2 b y' + a^2 y = 0$ where $b^2 < a^2$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]].
+Its [[Definition:Auxiliary Equation|auxiliary equation]] is:
+:$(2): \quad: m^2 + 2 b m + a^2 = 0$
+From [[Solution to Quadratic Equation with Real Coefficients]], the [[Definition:Root of Polynomial|roots]] of $(2)$ are:
+:$m_1 = - b + \sqrt {b^2 - a^2}$
+:$m_2 = - b - \sqrt {b^2 - a^2}$
+As $b^2 < a^2$, this converts into:
+:$m_1 = - b + i \sqrt {a^2 - b^2}$
+:$m_2 = - b - i \sqrt {a^2 - b^2}$
+These are [[Definition:Complex Number|complex]] and unequal.
+So from [[Solution of Constant Coefficient Homogeneous LSOODE]], the [[Definition:General Solution to Differential Equation|general solution]] of $(1)$ is:
+:$y = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$
+{{qed}}
+[[Category:Examples of Constant Coefficient Homogeneous LSOODEs]]
+9k5xhzn1dsruhtz976rp02pe76he2ci
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + 2 b y' + a^2 y = K cosine omega x/b less than a}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + 2 b y' + a^2 y = K \cos \omega x$ where $b^2 < a^2$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
+where:
+:$\alpha = \sqrt {a^2 - b^2}$
+:$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 2 b$
+:$q = a^2$
+:$\map R x = K \cos \omega x$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$y'' + 2 b y' + a^2 y = 0$
+From [[Linear Second Order ODE/y'' + 2 b y' + a^2 y = 0/b less than a|Linear Second Order ODE: $y'' + 2 b y' + a^2 y = 0: b < a$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$
+Substituting $\alpha = \sqrt {a^2 - b^2}$, this can be written more compactly as:
+:$y_g = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x}$
+We have that:
+:$\map R x = K \cos \omega x$
+It is noted that $K \cos \omega x$ is not itself a [[Definition:Particular Solution|particular solution]] of $(2)$.
+So from the [[Method of Undetermined Coefficients/Sine and Cosine|Method of Undetermined Coefficients for Sine and Cosine]]:
+:$y_p = A \sin \omega x + B \cos \omega x$
+where $A$ and $B$ are to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A \sin \omega x + B \cos \omega x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = \omega A \cos \omega x - \omega B \sin \omega x
+      | c = [[Derivative of Sine Function]], [[Derivative of Cosine Function]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = -\omega^2 A \sin \omega x - \omega^2 B \cos \omega x
+      | c = [[Power Rule for Derivatives]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = -\omega^2 \paren {A \sin \omega x + B \cos \omega x} + 2 b \omega \paren {A \cos \omega x - B \sin \omega x} + a^2 \paren {A \sin \omega x + B \cos \omega x}
+      | r = K \cos \omega x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\omega^2 A \sin \omega x - 2 b \omega B \sin \omega x + a^2 A \sin \omega x
+      | r = 0
+      | c = equating coefficients
+}}
+{{eqn | l = -\omega^2 B \cos \omega x + 2 b \omega A \cos \omega x + a^2 B \cos \omega x
+      | r = K \cos \omega x
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a^2 - \omega^2} A - 2 b \omega B
+      | r = 0
+      | c = 
+}}
+{{eqn | l = \paren {a^2 - \omega^2} B + 2 b \omega A
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {2 b \omega B} {a^2 - \omega^2}
+      | r = A
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \paren {a^2 - \omega^2} B + 2 b \omega \frac {2 b \omega B} {a^2 - \omega^2}
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = B \paren {\frac {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} {a^2 - \omega^2} }
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = K \frac {a^2 - \omega^2} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2}
+      | r = B
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {2 b \omega} {a^2 - \omega^2} K \frac {a^2 - \omega^2} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2}
+      | r = A
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = K \frac {2 b \omega} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2}
+      | r = A
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = K \frac {2 b \omega} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} \sin \omega x + K \frac {a^2 - \omega^2} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} \cos \omega x
+      | c = 
+}}
+{{eqn | r = \frac K {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} \paren {2 b \omega \sin \omega x + \paren {a^2 - \omega^2} \cos \omega x}
+      | c = 
+}}
+{{end-eqn}}
+From [[Multiple of Sine plus Multiple of Cosine/Cosine Form|Multiple of Sine plus Multiple of Cosine: Cosine Form]]:
+{{begin-eqn}}
+{{eqn | o = 
+      | r = 2 b \omega \sin \omega x + \paren {a^2 - \omega^2} \cos \omega x
+      | c = 
+}}
+{{eqn | r = \sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} \map \cos {\omega x + \map \arctan {\dfrac {-2 b \omega} {a^2 - \omega^2} } }
+      | c = 
+}}
+{{eqn | r = \sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} \map \cos {\omega x - \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} } }
+      | c = [[Tangent Function is Odd]]
+}}
+{{end-eqn}}
+and so:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = \frac {K \sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} \map \cos {\omega x - \phi}
+      | c = where $\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
+}}
+{{eqn | r = \frac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}
+      | c = 
+}}
+{{end-eqn}}
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
+where:
+:$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
+{{qed}}
+[[Category:Examples of Constant Coefficient LSOODEs]]
+[[Category:Examples of Method of Undetermined Coefficients]]
+8784espr5qejyaogjyeirfzv6k804c4
+\end{proof}<|endoftext|>
+\section{Condition for Resonance in Forced Vibration of Underdamped System}
+Tags: Mathematical Physics
+
+\begin{theorem}
+Consider a [[Definition:Physical System|physical system]] $S$ whose behaviour is defined by the [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = K \cos \omega x$
+where:
+:$K \in \R: k > 0$
+:$a, b \in \R_{>0}: b < a$
+Then $S$ is in [[Definition:Resonance|resonance]] when:
+:$\omega = \sqrt {a^2 - 2 b^2}$
+and thus the [[Definition:Resonance Frequency|resonance frequency]] is:
+:$\nu_R = \dfrac {\sqrt {a^2 - 2 b^2} } {2 \pi}$
+This [[Definition:Resonance Frequency|resonance frequency]] exists only when $a^2 - 2 b^2 > 0$.
+\end{theorem}
+
+\begin{proof}
+From [[Linear Second Order ODE/y'' + 2 b y' + a^2 y = K cosine omega x/b less than a|Linear Second Order ODE: $y'' + 2 b y' + a^2 y = K \cos \omega x$ where $b < a$]] the [[Definition:General Solution|general solution]] of $(1)$ is:
+:$(2): \quad y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
+where:
+:$\alpha = \sqrt {a^2 - b^2}$
+:$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
+Consider the [[Definition:Steady-State/Second Order ODE|steady-state]] component of $(2)$:
+:$y_s = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$
+By definition, the [[Definition:Amplitude of Simple Harmonic Motion|amplitude]] of $y_s$ is:
+:$A = \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} }$
+We have that both $a$ and $b$ are greater than $0$.
+It is also assumed that $\omega$ is also greater than $0$.
+Thus:
+:$4 b^2 \omega^2 > 0$
+:$a^2 > 0$
+:$\omega^2 > 0$
+Then $y_s$ is at a [[Definition:Maximum Element|maximum]] when:
+:$\map f \omega = 4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2$
+is at a [[Definition:Minimum Element|minimum]].
+[[Definition:Differentiation|Differentiating]] $\map f \omega$ {{WRT|Differentiation}} $\omega$:
+{{begin-eqn}}
+{{eqn | l = \map {f'} \omega
+      | r = 8 b^2 \omega + 2 \paren {a^2 - \omega^2} \paren {-2 \omega}
+      | c = 
+}}
+{{eqn | r = 8 b^2 \omega - 4 a^2 \omega + 4 \omega^3
+      | c = 
+}}
+{{eqn | r = 4 \omega \paren {2 b^2 - a^2 + \omega^2}
+      | c = 
+}}
+{{end-eqn}}
+Setting $\map {f'} \omega = 0$:
+{{begin-eqn}}
+{{eqn | l = 4 \omega \paren {2 b^2 - a^2 + \omega^2}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \omega
+      | r = 0
+      | c = 
+}}
+{{eqn | lo= \lor
+      | l = \omega^2
+      | r = a^2 - 2 b^2
+      | c = 
+}}
+{{end-eqn}}
+As $\omega > 0$ the first of these cannot apply.
+If $a^2 < 2 b^2$ then $a^2 - 2 b^2 < 0$ and so $\omega^2$ has no real roots.
+Hence the [[Definition:Resonance Frequency|resonance frequency]] is given by:
+:$\omega = \sqrt {a^2 - 2 b^2}$
+{{iff}} $a^2 > 2 b^2$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Filtered in Meet Semilattice}
+Tags: Join and Meet Semilattices, Upper Sets
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $H$ be a [[Definition:Empty Set|non-empty]] [[Definition:Upper Set|upper]] [[Definition:Subset|subset]] of $S$.
+Then $H$ is [[Definition:Filtered Subset|filtered]] {{iff}}
+:$\forall x, y \in H: x \wedge y \in H$
+\end{theorem}
+
+\begin{proof}
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]] of the proof of [[Directed in Join Semilattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Directed in Join Semilattice with Finite Suprema}
+Tags: Join and Meet Semilattices, Lower Sets
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
+Let $H$ be a [[Definition:Empty Set|non-empty]] [[Definition:Lower Set|lower]] [[Definition:Subset|subset]] of $S$.
+Then $H$ is [[Definition:Directed Subset|directed]] {{iff}}
+:for every [[Definition:Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $H$, $\sup A \in H$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let us assume that
+:$H$ is [[Definition:Directed Subset|directed]].
+Let $A$ be a [[Definition:Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $H$.
+By [[Directed iff Finite Subsets have Upper Bounds]]:
+:$\exists h \in H: \forall a \in A: a \preceq h$
+By definition
+:$z$ is [[Definition:Upper Bound of Set|upper bound]] of $A$
+By [[Existence of Non-Empty Finite Suprema in Join Semilattice]]:
+:$\sup A$ exists in $\left({S, \preceq}\right)$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup A \preceq h$
+Thus by definition of [[Definition:Lower Set|lower set]]:
+:$\sup A \in H$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let us assume that
+:for every [[Definition:Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $H$, $\sup A \in H$
+Let $x, y \in H$.
+:$A := \left\{ {x, y}\right\}$ is a [[Definition:Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $H$
+By assumption:
+:$\sup \left\{ {x, y}\right\} \in H$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup A$ is [[Definition:Upper Bound of Set|upper bound]] for $\left\{ {x, y}\right\}$
+Thus
+:$x \preceq \sup A \land y \preceq \sup A$
+Thus by definition
+:$H$ is [[Definition:Directed Subset|directed]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Resonance Frequency is less than Natural Frequency}
+Tags: Mathematical Physics
+
+\begin{theorem}
+Consider a [[Definition:Physical System|physical system]] $S$ whose behaviour is defined by the [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = K \cos \omega x$
+where:
+:$K \in \R: k > 0$
+:$a, b \in \R_{>0}: b < a$
+Then the [[Definition:Resonance Frequency|resonance frequency]] of $S$ is smaller than the [[Definition:Natural Frequency|natural frequency]] of the associated [[Definition:Second Order ODE|second order ODE]]:
+:$(2): \quad \dfrac {\d^2 y} {\d x^2} + 2 b \dfrac {\d y} {\d x} + a^2 x = 0$
+\end{theorem}
+
+\begin{proof}
+From [[Natural Frequency of Underdamped System]], the [[Definition:Natural Frequency|natural frequency]] of $(2)$ is:
+:$\nu = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$
+From [[Condition for Resonance in Forced Vibration of Underdamped System]], the [[Definition:Resonance Frequency|resonance frequency]] of $S$ is:
+:$\omega = \sqrt {a^2 - 2 b^2}$
+:$\nu_R = \dfrac {\sqrt {a^2 - 2 b^2} } {2 \pi}$
+We have that:
+{{begin-eqn}}
+{{eqn | l = \dfrac \nu {\nu_R}
+      | r = \dfrac {\sqrt {a^2 - b^2} } {\sqrt {a^2 - 2 b^2} }
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac {\nu^2} { {\nu_R}^2}
+      | r = \dfrac {a^2 - b^2} {a^2 - 2 b^2}
+      | c = 
+}}
+{{eqn | r = \dfrac {a^2 - 2 b^2 + b^2} {a^2 - 2 b^2}
+      | c = 
+}}
+{{eqn | r = 1 + \dfrac {b^2} {a^2 - 2 b^2}
+      | c = 
+}}
+{{end-eqn}}
+Thus $\nu_R < \nu$ {{iff}} $a^2 - 2 b^2 > 0$.
+But from [[Condition for Resonance in Forced Vibration of Underdamped System]], in order for there to be [[Definition:Resonance|resonance]], $a^2 - 2 b^2 > 0$.
+The two conditions are compatible, and so hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Filtered in Meet Semilattice with Finite Infima}
+Tags: Join and Meet Semilattices, Upper Sets
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $H$ be a [[Definition:Empty Set|non-empty]] [[Definition:Upper Set|lower]] [[Definition:Subset|subset]] of $S$.
+Then $H$ is [[Definition:Filtered Subset|filtered]] {{iff}}
+:for every [[Definition:Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $H$, $\inf A \in H$
+\end{theorem}
+
+\begin{proof}
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]] of the proof of [[Directed in Join Semilattice with Finite Suprema]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Interval between Local Maxima for Underdamped Free Vibration}
+Tags: Mathematical Physics
+
+\begin{theorem}
+Consider a [[Definition:Physical System|physical system]] $S$ whose behaviour can be described with the [[Definition:Second Order ODE|second order ODE]] in the form:
+:$(1): \quad \dfrac {\mathrm d^2 x} {\mathrm d t^2} + 2 b \dfrac {\mathrm d x} {\mathrm d t} + a^2 x = 0$
+for $a, b \in \R_{>0}$.
+Let $b < a$, so as to make $S$ [[Definition:Underdamped|underdamped]].
+:[[File:UnderdampedPeriodAmplitude.png|600px]]
+Let $T$ be the [[Definition:Period of Underdamped Oscillation|period of oscillation]] of $S$.
+Then the successive [[Definition:Local Maximum|local maxima]] of $x$ occur for $t = 0, T, 2T, \ldots$
+\end{theorem}
+
+\begin{proof}
+Let the position of $S$ be described in the [[Canonical Form of Underdamped Oscillatory System|canonical form]]:
+:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \cos \left({\alpha t - \theta}\right)$
+where:
+:$\alpha = \sqrt {a^2 - b^2}$.
+:$\theta = \arctan \left({\dfrac b \alpha}\right)$
+From [[Period of Oscillation of Underdamped System is Regular]], the [[Definition:Period of Underdamped Oscillation|period of oscillation]] $T$ is given by:
+:$T = \dfrac {2 \pi} {a^2 - b^2}$
+[[Definition:Differentiation|Differentiating]] {{{WRT}} $t$:
+{{begin-eqn}}
+{{eqn | l = x'
+      | r = -b \dfrac {x_0 \, a} \alpha e^{-b t} \cos \left({\alpha t - \theta}\right) - \dfrac {x_0 \, a} \alpha e^{-b t} \alpha \sin \left({\omega t - \theta}\right)
+      | c = 
+}}
+{{eqn | r = -\dfrac {x_0 \, a} \alpha e^{-b t} \left({b \cos \left({\alpha t - \theta}\right) + \alpha \sin \left({\alpha t - \theta}\right)}\right)
+      | c = 
+}}
+{{end-eqn}}
+From [[Derivative at Maximum or Minimum]], the [[Definition:Local Maximum|local maxima]] and [[Definition:Local Minimum|local minima]] of $x$ occur at $x' = 0$:
+{{begin-eqn}}
+{{eqn | l = x'
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = -\dfrac {x_0 \, a} \alpha e^{-b t} \left({b \cos \left({\alpha t - \theta}\right) + \alpha \sin \left({\alpha t - \theta}\right)}\right)
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = b \cos \left({\alpha t - \theta}\right)
+      | r = -\alpha \sin \left({\alpha t - \theta}\right)
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tan \left({\alpha t - \theta}\right)
+      | r = -\frac b \alpha
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\tan \alpha t - \tan \theta} {1 + \tan \alpha t \tan \theta}
+      | r = -\frac b \alpha
+      | c = [[Tangent of Difference]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\tan \alpha t - \dfrac b \alpha} {1 + \tan \alpha t \dfrac b \alpha}
+      | r = -\frac b \alpha
+      | c = Definition of $\theta$
+}}
+{{eqn | ll= \leadsto
+      | l = \tan \alpha t - \dfrac b \alpha 
+      | r = -\frac b \alpha - \tan \alpha t \dfrac {b^2} {\alpha^2}
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tan \alpha t \left({1 + \dfrac {b^2} {\alpha^2} }\right)
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \tan \alpha t
+      | r = 0
+      | c = as $1 + \dfrac {b^2} {\alpha^2} > 0$
+}}
+{{eqn | ll= \leadsto
+      | l = \alpha t
+      | r = 0, \pi, 2 \pi, \ldots
+      | c = [[Tangent of Zero]], [[Tangent of Angle plus Straight Angle]]
+}}
+{{end-eqn}}
+It remains to be determined which of these points at which $x' = 0$ are [[Definition:Local Maximum|local maxima]].
+This occurs when $x > 0$.
+From [[Cosine of Angle plus Full Angle]]:
+:$\cos \left({x}\right) = \cos \left({2 \pi + x}\right)$
+We have that at $x = x_0$ at $t = 0$.
+It is given that $x_0 > 0$.
+So at $t = 0, 2 \pi, 4 \pi, \ldots$ we have that:
+:$\cos \alpha t > 0$
+Similarly, from [[Cosine of Angle plus Straight Angle]]:
+:$\cos \left({x}\right) = -\cos \left({\pi + x}\right)$
+So at $t = \pi, 3 \pi, 5 \pi, \ldots$ we have that:
+:$\cos \alpha t < 0$
+Thus we have that:
+:$\alpha L = 2 \pi$
+where $L$ is the value of $t$ between consecutive [[Definition:Local Maximum|local maxima]] of $x$.
+Thus:
+:$L = \dfrac {2 \pi} {\alpha} = \dfrac {2 \pi} {a^2 - b^2} = T$
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Ratio of Successive Local Maxima for Underdamped Free Vibration}
+Tags: Mathematical Physics
+
+\begin{theorem}
+Consider a [[Definition:Physical System|physical system]] $S$ whose behaviour can be described with the [[Definition:Second Order ODE|second order ODE]] in the form:
+:$(1): \quad \dfrac {\mathrm d^2 x} {\mathrm d t^2} + 2 b \dfrac {\mathrm d x} {\mathrm d t} + a^2 x = 0$
+for $a, b \in \R_{>0}$.
+Let $b < a$, so as to make $S$ [[Definition:Underdamped|underdamped]].
+:[[File:UnderdampedPeriodAmplitude.png|600px]]
+Let $T$ be the [[Definition:Period of Underdamped Oscillation|period of oscillation]] of $S$.
+Let $x_1$ and $x_2$ be successive [[Definition:Local Maximum|local maxima]] of $x$.
+Then:
+:$\dfrac {x_1} {x_2} = e^{b T}$
+\end{theorem}
+
+\begin{proof}
+Let the position of $S$ be described in the [[Canonical Form of Underdamped Oscillatory System|canonical form]]:
+:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \cos \left({\alpha t - \theta}\right)$
+where:
+:$\alpha = \sqrt {a^2 - b^2}$.
+:$\theta = \arctan \left({\dfrac b \alpha}\right)$
+From [[Period of Oscillation of Underdamped System is Regular]], the [[Definition:Period of Underdamped Oscillation|period of oscillation]] $T$ is given by:
+:$T = \dfrac {2 \pi} {a^2 - b^2}$
+From [[Interval between Local Maxima for Underdamped Free Vibration]], the [[Definition:Local Maximum|local maxima]] of $x$ occur at:
+:$t = 0, T, 2 T, \ldots$
+Let $x_1$ occur at $n T$.
+Thus $x_2$ occurs at $\left({n + 1}\right) T$.
+[[Definition:Differentiation|Differentiating]] {{{WRT}} $t$:
+{{begin-eqn}}
+{{eqn | l = x_1
+      | r = \dfrac {x_0 \, a} \alpha e^{-b n T} \cos \left({\alpha n T - \theta}\right)
+      | c = 
+}}
+{{eqn | l = x_2
+      | r = \dfrac {x_0 \, a} \alpha e^{-b \left({n + 1}\right) T} \cos \left({\alpha \left({n + 1}\right) T - \theta}\right)
+      | c = 
+}}
+{{end-eqn}}
+As $T = \dfrac {2 \pi} {a^2 - b^2} = \dfrac {2 \pi} \alpha$, we have:
+{{begin-eqn}}
+{{eqn | l = \cos \left({\alpha \left({n + 1}\right) T - \theta}\right)
+      | r = \cos \left({\alpha \left({n + 1}\right) \dfrac {2 \pi} \alpha - \theta}\right)
+      | c = 
+}}
+{{eqn | r = \cos \left({\left({n + 1}\right) 2 \pi - \theta}\right)
+      | c = 
+}}
+{{eqn | r = \cos \left({n 2 \pi - \theta}\right)
+      | c = 
+}}
+{{eqn | r = \cos \left({\alpha n \dfrac {2 \pi} \alpha  - \theta}\right)
+      | c = 
+}}
+{{eqn | r = \cos \left({\alpha n T  - \theta}\right)
+      | c = 
+}}
+{{end-eqn}}
+Hence:
+{{begin-eqn}}
+{{eqn | l = \dfrac {x_1} {x_2}
+      | r = \dfrac {\dfrac {x_0 \, a} \alpha e^{-b n T} \cos \left({\alpha n T - \theta}\right)} {\dfrac {x_0 \, a} \alpha e^{-b \left({n + 1}\right) T} \cos \left({\alpha \left({n + 1}\right) T - \theta}\right)}
+      | c = 
+}}
+{{eqn | r = \dfrac {e^{-b n T} \cos \left({\alpha n T - \theta}\right)} {e^{-b \left({n + 1}\right) T} \cos \left({\alpha n T - \theta}\right)}
+      | c = 
+}}
+{{eqn | r = \dfrac {e^{-b n T} } {e^{-b \left({n + 1}\right) T} }
+      | c = 
+}}
+{{eqn | r = e^{-b n T + b \left({n + 1}\right) T}
+      | c = 
+}}
+{{eqn | r = e^{b T}
+      | c = 
+}}
+{{end-eqn}}
+as required.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Motion of Particle in Polar Coordinates}
+Tags: Mathematical Physics
+
+\begin{theorem}
+Consider a [[Definition:Particle|particle]] $p$ of [[Definition:Mass|mass]] $m$ moving in the [[Definition:Plane|plane]] under the influence of a [[Definition:Force|force]] $\mathbf F$.
+Let the [[Definition:Position|position]] of $p$ at [[Definition:Time|time]] $t$ be given in [[Definition:Polar Coordinates|polar coordinates]] as $\left\langle{r, \theta}\right\rangle$.
+Let $\mathbf F$ be expressed as:
+:$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$
+where:
+:$\mathbf u_r$ is the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Radial Coordinate|radial coordinate]] of $p$
+:$\mathbf u_\theta$ is the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Angular Coordinate|angular coordinate]] of $p$
+:$F_r$ and $F_\theta$ are the [[Definition:Magnitude|magnitudes]] of the components of $\mathbf F$ in the directions of $\mathbf u_r$ and $\mathbf u_\theta$ respectively.
+Then the [[Definition:Second Order ODE|second order ordinary differential equations]] governing the motion of $m$ under the [[Definition:Force|force]] $\mathbf F$ are:
+{{begin-eqn}}
+{{eqn | l = F_\theta
+      | r = m \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right)
+      | c = 
+}}
+{{eqn | l = F_r
+      | r = m \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right)
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Radius Vector|radius vector]] $\mathbf r$ from the [[Definition:Origin|origin]] to $p$ be expressed as:
+:$(1): \quad \mathbf r = r \mathbf u_r$
+From [[Velocity Vector in Polar Coordinates]], the [[Definition:Velocity|velocity]] $\mathbf v$ of $p$ can be expressed in [[Definition:Vector|vector]] form as:
+:$\mathbf v = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r$
+From [[Acceleration Vector in Polar Coordinates]], the the [[Definition:Acceleration|acceleration]] $\mathbf a$ of $p$ can be expressed as:
+:$\mathbf a = \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r$
+We have:
+:$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$
+and from [[Newton's Second Law of Motion]]:
+:$\mathbf F = m \mathbf a$
+Hence:
+:$\mathbf F = m \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + m \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r$
+Equating components:
+{{begin-eqn}}
+{{eqn | l = F_r \mathbf u_r
+      | r = m \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r
+      | c = 
+}}
+{{eqn | l = F_\theta \mathbf u_\theta
+      | r =  m \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta
+      | c = 
+}}
+{{end-eqn}}
+Hence the result:
+{{begin-eqn}}
+{{eqn | l = F_\theta
+      | r =  m \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right)
+      | c = 
+}}
+{{eqn | l = F_r
+      | r = m \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivative of Angular Component under Central Force}
+Tags: Mechanics
+
+\begin{theorem}
+Let a [[Definition:Particle|point mass]] $p$ of [[Definition:Mass|mass]] $m$ be under the influence of a [[Definition:Central Force|central force]] $\mathbf F$.
+Let the [[Definition:Position|position]] of $p$ at [[Definition:Time|time]] $t$ be given in [[Definition:Polar Coordinates|polar coordinates]] as $\polar {r, \theta}$.
+Let $\mathbf r$ be the [[Definition:Radius Vector|radius vector]] from the [[Definition:Origin|origin]] to $p$.
+Then the [[Definition:Rate of Change|rate of change]] of the [[Definition:Angular Coordinate|angular coordinate]] of $p$ is [[Definition:Inverse Proportion|inversely proportional]] to the [[Definition:Square (Algebra)|square]] of the [[Definition:Radial Coordinate|radial coordinate]] of $p$.
+\end{theorem}
+
+\begin{proof}
+Let $\mathbf F$ be expressed as:
+:$\mathbf F = F_r \mathbf u_r + F_\theta \mathbf u_\theta$
+where:
+:$\mathbf u_r$ is the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Radial Coordinate|radial coordinate]] of $p$
+:$\mathbf u_\theta$ is the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Angular Coordinate|angular coordinate]] of $p$
+:$F_r$ and $F_\theta$ are the [[Definition:Magnitude|magnitudes]] of the components of $\mathbf F$ in the directions of $\mathbf u_r$ and $\mathbf u_\theta$ respectively.
+From [[Motion of Particle in Polar Coordinates]], the [[Definition:Second Order ODE|second order ordinary differential equations]] governing the motion of $m$ under the [[Definition:Force|force]] $\mathbf F$ are:
+{{begin-eqn}}
+{{eqn | l = F_\theta
+      | r = m \paren {r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t} }
+      | c = 
+}}
+{{eqn | l = F_r
+      | r = m \paren {\dfrac {\d^2 r} {\d t^2} - r \paren {\dfrac {\d \theta} {\d t} }^2}
+      | c = 
+}}
+{{end-eqn}}
+However, we are given that $\mathbf F$ is a [[Definition:Central Force|central force]].
+:[[File:CentralForce.png|600px]]
+Thus:
+{{begin-eqn}}
+{{eqn | l = F_\theta
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r \dfrac {\d^2 \theta} {\d t^2} + 2 \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t}
+      | r = 0
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = r^2 \dfrac {\d^2 \theta} {\d t^2} + 2 r \dfrac {\d r} {\d t} \dfrac {\d \theta} {\d t}
+      | r = 0
+      | c = multiplying through by $r$
+}}
+{{eqn | ll= \leadsto
+      | l = \dfrac \d {\d t} \paren {r^2 \dfrac {\d \theta} {\d t} }
+      | r = 0
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | ll= \leadsto
+      | l = r^2 \dfrac {\d \theta} {\d t}
+      | r = h
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+for some [[Definition:Constant|constant]] $h$.
+That is:
+:$\dfrac {\d \theta} {\d t} = \dfrac h {r^2}$
+Hence the result, by definition of [[Definition:Inverse Proportion|inverse proportion]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Linear Second Order ODE/y'' + y = K}
+Tags: Examples of Constant Coefficient LSOODEs, Examples of Method of Undetermined Coefficients
+
+\begin{theorem}
+The [[Definition:Second Order ODE|second order ODE]]:
+:$(1): \quad y'' + y = K$
+has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y = C_1 \sin x + C_2 \cos x + K$
+\end{theorem}
+
+\begin{proof}
+It can be seen that $(1)$ is a [[Definition:Nonhomogeneous Linear Second Order ODE|nonhomogeneous linear second order ODE]] with [[Definition:Constant|constant]] [[Definition:Coefficient|coefficients]] in the form:
+:$y'' + p y' + q y = \map R x$
+where:
+:$p = 0$
+:$q = 1$
+:$\map R x = K$
+First we establish the solution of the corresponding [[Definition:Constant Coefficient Homogeneous Linear Second Order ODE|constant coefficient homogeneous linear second order ODE]]:
+:$(2): \quad y'' + y = 0$
+From [[Linear Second Order ODE/y'' + y = 0|Linear Second Order ODE: $y'' + y = 0$]], this has the [[Definition:General Solution to Differential Equation|general solution]]:
+:$y_g = C_1 \sin x + C_2 \cos x$
+We have that:
+:$\map R x = K$
+So from the [[Method of Undetermined Coefficients/Polynomial|Method of Undetermined Coefficients for Polynomial]]:
+:$y_p = A K$
+where $A$ is to be determined.
+Hence:
+{{begin-eqn}}
+{{eqn | l = y_p
+      | r = A K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}'
+      | r = 0
+      | c = [[Derivative of Constant]]
+}}
+{{eqn | ll= \leadsto
+      | l = {y_p}''
+      | r = 0
+      | c = [[Derivative of Constant]]
+}}
+{{end-eqn}}
+Substituting into $(1)$:
+{{begin-eqn}}
+{{eqn | l = A K
+      | r = K
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = A
+      | r = 1
+}}
+{{end-eqn}}
+Our work is done.
+So from [[General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution]]:
+:$y = y_g + y_p = C_1 \sin x + C_2 \cos x + K$
+{{qed}}
+[[Category:Examples of Constant Coefficient LSOODEs]]
+[[Category:Examples of Method of Undetermined Coefficients]]
+qowaor0rptc3den7cksl9o1zijqpv4k
+\end{proof}<|endoftext|>
+\section{Velocity Vector in Polar Coordinates}
+Tags: Polar Coordinates
+
+\begin{theorem}
+Consider a [[Definition:Particle|particle]] $p$ moving in the [[Definition:Plane|plane]].
+Let the [[Definition:Position|position]] of $p$ at [[Definition:Time|time]] $t$ be given in [[Definition:Polar Coordinates|polar coordinates]] as $\left\langle{r, \theta}\right\rangle$.
+Then the [[Definition:Velocity|velocity]] $\mathbf v$ of $p$ can be expressed as:
+:$\mathbf v = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r$
+where:
+:$\mathbf u_r$ is the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Radial Coordinate|radial coordinate]] of $p$
+:$\mathbf u_\theta$ is the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Angular Coordinate|angular coordinate]] of $p$
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Radius Vector|radius vector]] $\mathbf r$ from the [[Definition:Origin|origin]] to $p$ be expressed as:
+:$(1): \quad \mathbf r = r \mathbf u_r$
+:[[File:MotionInPolarPlane.png|600px]]
+From [[Derivatives of Unit Vectors in Polar Coordinates]]:
+{{begin-eqn}}
+{{eqn | n = 2
+      | l = \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta}
+      | r = \mathbf u_\theta
+      | c = 
+}}
+{{eqn | n = 3
+      | l = \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta}
+      | r = -\mathbf u_r
+      | c = 
+}}
+{{end-eqn}}
+The [[Definition:Velocity|velocity]] of $p$ is by definition the [[Definition:Rate of Change|rate of change]] in its [[Definition:Position|position]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf v
+      | r = \dfrac {\mathrm d \mathbf r} {\mathrm d t}
+      | c = 
+}}
+{{eqn | r = r \dfrac {\mathrm d \mathbf u_r} {\mathrm d t} + \mathbf u_r \dfrac {\mathrm d r} {\mathrm d t}
+      | c = from $(1)$ and [[Product Rule for Derivatives]]
+}}
+{{eqn | r = r \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta} \dfrac {\mathrm d \theta} {\mathrm d t} + \mathbf u_r \dfrac {\mathrm d r} {\mathrm d t}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r
+      | c = substituting from $(2)$ and $(3)$
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Derivatives of Unit Vectors in Polar Coordinates}
+Tags: Polar Coordinates
+
+\begin{theorem}
+Consider a [[Definition:Particle|particle]] $p$ moving in the [[Definition:Plane|plane]].
+Let the [[Definition:Position|position]] of $p$ be given in [[Definition:Polar Coordinates|polar coordinates]] as $\left\langle{r, \theta}\right\rangle$.
+Let:
+:$\mathbf u_r$ be the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Radial Coordinate|radial coordinate]] of $p$
+:$\mathbf u_\theta$ be the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Angular Coordinate|angular coordinate]] of $p$
+Then the [[Definition:Derivative|derivative]] of $\mathbf u_r$ and $\mathbf u_\theta$ {{WRT}} $\theta$ can be expressed as:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta}
+      | r = \mathbf u_\theta
+}}
+{{eqn | l = \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta}
+      | r = -\mathbf u_r
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+By definition of [[Definition:Sine|sine]] and [[Definition:Cosine|cosine]]:
+{{begin-eqn}}
+{{eqn | n = 1
+      | l = \mathbf u_r
+      | r = \mathbf i \cos \theta + \mathbf j \sin \theta
+}}
+{{eqn | n = 2
+      | l = \mathbf u_\theta
+      | r = -\mathbf i \sin \theta + \mathbf j \cos \theta
+}}
+{{end-eqn}}
+where $\mathbf i$ and $\mathbf j$ are the [[Definition:Unit Vector|unit vectors]] in the [[Definition:X-Axis|$x$-axis]] and [[Definition:Y-Axis|$y$-axis]] respectively.
+:[[File:DerivativesOfUnitPolarVectors.png|600px]]
+[[Definition:Differentiation|Differentiating]] $(1)$ and $(2)$ {{WRT}} $\theta$ gives:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta}
+      | r = -\mathbf i \sin \theta + \mathbf j \cos \theta
+      | c = 
+}}
+{{eqn | r = \mathbf u_\theta
+      | c = 
+}}
+{{eqn | l = \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta}
+      | r = -\mathbf i \cos \theta - \mathbf j \sin \theta
+      | c = 
+}}
+{{eqn | r = -\mathbf u_r
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Acceleration Vector in Polar Coordinates}
+Tags: Polar Coordinates
+
+\begin{theorem}
+Consider a [[Definition:Particle|particle]] $p$ moving in the [[Definition:Plane|plane]].
+Let the [[Definition:Position|position]] of $p$ at [[Definition:Time|time]] $t$ be given in [[Definition:Polar Coordinates|polar coordinates]] as $\left\langle{r, \theta}\right\rangle$.
+Then the [[Definition:Acceleration|acceleration]] $\mathbf a$ of $p$ can be expressed as:
+:$\mathbf a = \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r$
+where:
+:$\mathbf u_r$ is the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Radial Coordinate|radial coordinate]] of $p$
+:$\mathbf u_\theta$ is the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Angular Coordinate|angular coordinate]] of $p$
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Radius Vector|radius vector]] $\mathbf r$ from the [[Definition:Origin|origin]] to $p$ be expressed as:
+:$(1): \quad \mathbf r = r \mathbf u_r$
+:[[File:MotionInPolarPlane.png|600px]]
+From [[Derivatives of Unit Vectors in Polar Coordinates]]:
+{{begin-eqn}}
+{{eqn | n = 2
+      | l = \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta}
+      | r = \mathbf u_\theta
+      | c = 
+}}
+{{eqn | n = 3
+      | l = \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta}
+      | r = -\mathbf u_r
+      | c = 
+}}
+{{end-eqn}}
+From [[Velocity Vector in Polar Coordinates]]:
+:$\mathbf v = r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_r$
+where $\mathbf v$ is the [[Definition:Velocity|velocity]] of $p$.
+The [[Definition:Acceleration|acceleration]] of $p$ is by definition the [[Definition:Rate of Change|rate of change]] in its [[Definition:Velocity|velocity]]:
+{{begin-eqn}}
+{{eqn | l = \mathbf a
+      | r = \dfrac {\mathrm d \mathbf v} {\mathrm d t}
+      | c = 
+}}
+{{eqn | r = r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + r \dfrac {\mathrm d \theta} {\mathrm d t} \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \mathbf u_r} {\mathrm d t}
+      | c = [[Product Rule for Derivatives]]
+}}
+{{eqn | r = r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta + r \dfrac {\mathrm d \theta} {\mathrm d t} \dfrac {\mathrm d \mathbf u_\theta} {\mathrm d \theta} \dfrac {\mathrm d \theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \mathbf u_r} {\mathrm d \theta} \dfrac {\mathrm d \theta} {\mathrm d t}
+      | c = [[Chain Rule for Derivatives]]
+}}
+{{eqn | r = r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} \mathbf u_\theta + \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_\theta - r \dfrac {\mathrm d \theta} {\mathrm d t} \mathbf u_r \dfrac {\mathrm d \theta} {\mathrm d t} + \dfrac {\mathrm d^2 r} {\mathrm d t^2} \mathbf u_r + \dfrac {\mathrm d r} {\mathrm d t} \mathbf u_\theta \dfrac {\mathrm d \theta} {\mathrm d t}
+      | c = substituting from $(2)$ and $(3)$
+}}
+{{eqn | r = \left({r \dfrac {\mathrm d^2 \theta} {\mathrm d t^2} + 2 \dfrac {\mathrm d r} {\mathrm d t} \dfrac {\mathrm d \theta} {\mathrm d t} }\right) \mathbf u_\theta + \left({\dfrac {\mathrm d^2 r} {\mathrm d t^2} - r \left({\dfrac {\mathrm d \theta} {\mathrm d t} }\right)^2}\right) \mathbf u_r
+      | c = gathering terms
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Eccentricity of Orbit indicates its Total Energy}
+Tags: Celestial Mechanics
+
+\begin{theorem}
+Consider a [[Definition:Planet|planet]] $p$ of [[Definition:Mass|mass]] $m$ orbiting a [[Definition:Star (Physics)|star]] $S$ of [[Definition:Mass|mass]] $M$ under the influence of the [[Definition:Gravitational Field|gravitational field]] which the two [[Definition:Body|bodies]] give rise to.
+Then the total [[Definition:Energy|energy]] of the system determines the [[Definition:Eccentricity of Conic Section|eccentricity]] of the [[Definition:Orbit (Physics)|orbit]] of $p$ around $S$.
+\end{theorem}
+
+\begin{proof}
+Let:
+:$\mathbf u_r$ be the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Radial Coordinate|radial coordinate]] of $p$
+:$\mathbf u_\theta$ be the [[Definition:Unit Vector|unit vector]] in the direction of the [[Definition:Angular Coordinate|angular coordinate]] of $p$.
+By [[Kinetic Energy of Motion]], the [[Definition:Kinetic Energy|kinetic energy]] of $p$ is:
+:$K = \dfrac {m v^2} 2$
+where $v$ is the [[Definition:Magnitude|magnitude]] of the [[Definition:Velocity|velocity]] of $p$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = K
+      | r = \dfrac {m \mathbf v \cdot \mathbf v} 2
+      | c = [[Dot Product of Vector with Itself]]
+}}
+{{eqn | r = \dfrac 1 2 m \paren {r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r} \cdot \paren {r \dfrac {\d \theta} {\d t} \mathbf u_\theta + \dfrac {\d r} {\d t} \mathbf u_r}
+      | c = [[Velocity Vector in Polar Coordinates]]
+}}
+{{eqn | n = 1
+      | r = \dfrac 1 2 m \paren {r^2 \paren {\dfrac {\d \theta} {\d t} }^2 + \paren {\dfrac {\d r} {\d t} }^2}
+      | c = {{Defof|Dot Product}}
+}}
+{{end-eqn}}
+The [[Definition:Potential Energy|potential energy]] $P$ of the system is the negative of the [[Definition:Work|work]] required to move $p$ to [[Definition:Infinity|infinity]]:
+{{MissingLinks|Definition of Work, the above definition of P.E.}}
+{{begin-eqn}}
+{{eqn | l = P
+      | r = -\int_r^\infty \dfrac {G M m} {r^2} \rd r
+      | c = [[Newton's Law of Universal Gravitation]]
+}}
+{{eqn | r = \intlimits {\dfrac {G M m} r} r \infty
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = -\dfrac {G M m} r
+      | c = 
+}}
+{{eqn | n = 2
+      | r = -\dfrac {k m} r
+      | c = defining $k = G M$: constant for this system
+}}
+{{end-eqn}}
+By the [[Principle of Conservation of Energy]], the total [[Definition:Energy|energy]] in the system remains [[Definition:Constant|constant]]: $E$, say.
+So:
+:$E = \dfrac 1 2 m \paren {r^2 \paren {\dfrac {\d \theta} {\d t} }^2 + \paren {\dfrac {\d r} {\d t} }^2} - \dfrac {k m} r$
+{{WLOG}}, let us arrange the [[Definition:Polar Axis|polar axis]] so as to make $r$ a [[Definition:Local Minimum|minimum]] when $\theta = 0$.
+By [[Kepler's First Law of Planetary Motion]], the position of $p$ in [[Definition:Polar Coordinates|polar coordinates]] is:
+:$(3): \quad r = \dfrac {h^2 / k} {1 + e \cos \theta}$
+At the instant when $\theta = 0$, we therefore have:
+:$r = \dfrac {h^2 / k} {1 + e}$
+At this point, $r$ is at a [[Definition:Local Minimum|local minimum]].
+Hence:
+:$\dfrac {m r^2} 2 \dfrac {h^2} {r^4} - \dfrac {k m} r = E$
+Eliminating $r$ from these gives:
+:$e = \sqrt {1 + E \paren {\dfrac {2 h^2} {m k^2} } }$
+Thus equation $(3)$ for the [[Definition:Orbit (Physics)|orbit]] of $p$ can be written as:
+:$r = \dfrac {h^2 / k} {1 + \sqrt {1 + E \paren {2 h^2 / m k^2} \cos \theta} }$
+Thus from [[Equation of Conic Section in Polar Form]], it can be seen that the [[Definition:Orbit (Physics)|orbit]] is:
+:an [[Definition:Ellipse|ellipse]] when $E < 0$
+:a [[Definition:Parabola|parabola]] when $E = 0$
+:a [[Definition:Hyperbola|hyperbola]] when $E > 0$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Non-Empty Finite Suprema in Join Semilattice}
+Tags: Join and Meet Semilattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Join Semilattice|join semilattice]].
+Let $A$ be a [[Definition:Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$.
+Then $A$ admits a [[Definition:Supremum of Set|supremum]] in $\left({S, \preceq}\right)$.
+\end{theorem}
+
+\begin{proof}
+We will prove by [[Principle of Mathematical Induction|induction]] of the [[Definition:Cardinality|cardinality]] of [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $H$.
+=== Base case ===
+:$\forall A \subseteq S: \left\vert{A}\right\vert = 1 \implies \exists \sup A$
+where $\left\vert{A}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $A$.
+Let $A \subseteq S$ such that
+:$\left\vert{A}\right\vert = 1$
+By [[Cardinality of Singleton]]:
+:$\exists a: A = \left\{ {a}\right\}$
+By definition of [[Definition:Subset|subset]]:
+:$a \in S$
+Thus by [[Supremum of Singleton]]:
+:the [[Definition:Supremum of Set|supremum]] of $A$ exists in $\left({S, \preceq}\right)$.
+=== Induction Hypothesis  ===
+:$n \ge 1 \land \forall A \subseteq S: \left\vert{A}\right\vert = n \implies \exists \sup A$
+=== Induction Step ===
+:$\forall A \subseteq S: \left\vert{A}\right\vert = n+1 \implies \exists \sup A$
+Let $A \subseteq S$ such that
+:$\left\vert{A}\right\vert = n+1$
+By definition of [[Definition:Cardinality of Finite Set|cardinality of finite set]]:
+:$A \sim \N_{< n+1}$
+where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]].
+By definition of [[Definition:Set Equivalence|set equivalence]]:
+:there exists a [[Definition:Bijection|bijection]] $f: \N_{< n+1} \to A$
+By [[Restriction of Injection is Injection]]:
+:$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is an [[Definition:Injection|injection]].
+By definition
+:$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is a [[Definition:Surjection|surjection]].
+By definition
+:$f \restriction_{\N_{< n}}: \N_{< n} \to f^\to\left({\N_{< n}}\right)$ is a [[Definition:Bijection|bijection]].
+By definition of [[Definition:Set Equivalence|set equivalence]]:
+:$\N_{< n} \sim f^\to\left({\N_{< n}}\right)$
+By definition of [[Definition:Cardinality of Finite Set|cardinality of finite set]]:
+:$\left\vert{f^\to\left({\N_{< n}}\right)}\right\vert = n$
+By definitions of [[Definition:Image of Subset under Mapping|image of set]] and [[Definition:Subset|subset]]:
+:$f^\to\left({\N_{< n}}\right) \subseteq A$
+By [[Subset Relation is Transitive]]:
+:$f^\to\left({\N_{< n}}\right) \subseteq S$
+By Induction Hypothesis:
+:$\exists \sup \left({ f^\to\left({\N_{< n}}\right)}\right)$
+By definition $\N_{< n+1}$
+:$n \in \N_{< n+1}$
+By definition of [[Definition:Mapping|mapping]]:
+:$f \left({n}\right) \in A$
+By definition of [[Definition:Subset|subset]]:
+:$f \left({n}\right) \in S$
+{{begin-eqn}}
+{{eqn |l = A
+      |r = f^\to\left({\N_{< n+1} }\right)
+| c = definition of [[Definition:Surjection|surjection]]
+}}
+{{eqn |r =  f^\to\left({\N_{< n}\cup \left\{ {n}\right\} }\right)
+| c = definition of [[Definition:Initial Segment of Natural Numbers|initial segment of natural numbers]]
+}}
+{{eqn |r = f^\to\left({\N_{< n} }\right) \cup f^\to\left({\left\{ {n}\right\} }\right)
+| c =[[Image of Union under Mapping]]
+}}
+{{eqn |r = f^\to\left({\N_{< n} }\right) \cup \left\{ {f\left({n }\right)}\right\}
+| c =[[Image of Singleton under Mapping]]
+}}
+{{end-eqn}}
+Thus:
+{{begin-eqn}}
+{{eqn | l =\sup A
+      | r = \sup\left\{ {\sup \left({f^\to\left({\N_{< n} }\right)}\right), \sup \left\{ {f\left(n\right)}\right\} }\right\}
+      | c = [[Supremum of Suprema]]
+}}
+{{eqn | r = \sup\left\{ {\sup \left({f^\to\left({\N_{< n} }\right)}\right), f\left(n\right)}\right\}
+      | c = [[Supremum of Singleton]]
+}}
+{{eqn | r = \sup \left({f^\to\left({\N_{< n} }\right)}\right) \vee f\left(n\right)
+      | c = definition of [[Definition:Join (Order Theory)|join operation]]
+}}
+{{end-eqn}}
+Thus $A$ admits a [[Definition:Supremum of Set|supremum]] in $\left({S, preceq}\right)$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Product form of Riemann Zeta Function}
+Tags: Number Theory
+
+\begin{theorem}
+Let $s \in \R: s > 1$.
+Then:
+:$\displaystyle \sum_{k \mathop \in \N_{>0} } \dfrac 1 {k^s} = \prod_{p \mathop \in \Bbb P} \dfrac 1 {1 - 1 / p^s}$
+where $\Bbb P$ denotes the set of all [[Definition:Prime Number|prime numbers]].
+\end{theorem}<|endoftext|>
+\section{Variance of Exponential Distribution}
+Tags: Exponential Distribution, Variance, Variance of Exponential Distribution
+
+\begin{theorem}
+Let $X$ be a [[Definition:Continuous Random Variable|continuous random variable]] with the [[Definition:Exponential Distribution|exponential distribution]] with parameter $\beta$.
+Then the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is:
+:$\var X = \beta^2$
+\end{theorem}
+
+\begin{proof}
+From [[Variance as Expectation of Square minus Square of Expectation]]:
+:$\var X = \expect {X^2} - \paren {\expect X}^2$
+From [[Expectation of Exponential Distribution]]:
+:$\expect X = \beta$
+The expectation of $X^2$ is:
+{{begin-eqn}}
+{{eqn | l = \expect {X^2}
+      | r = \int_{x \mathop \in \Omega_X} x^2 \, \map {f_X} x \rd x
+      | c = {{Defof|Expectation of Continuous Random Variable}}
+}}
+{{eqn | r = \int_0^\infty x^2 \frac 1 \beta \map \exp {-\frac x \beta} \rd x
+      | c = [[PDF of Exponential Distribution]]
+}}
+{{eqn | r = \intlimits {-x^2 \map \exp {-\frac x \beta} } 0 \infty + \int_0^\infty 2 x \map \exp {-\frac x \beta} \rd x
+      | c = [[Integration by Parts]]
+}}
+{{eqn | r = 0 + 2 \beta \int_0^\infty x \frac 1 \beta \, \map \exp {-\frac x \beta} \rd x 
+      | c = algebraic manipulation
+}}
+{{eqn | r = 2 \beta \, \expect X
+      | c = [[Expectation of Exponential Distribution]]
+}}
+{{eqn | r = 2 \beta^2
+}}
+{{end-eqn}}
+Thus the [[Definition:Variance of Discrete Random Variable|variance]] of $X$ is:
+{{begin-eqn}}
+{{eqn | l = \var X
+      | r = \expect {X^2} - \paren {\expect X}^2
+}}
+{{eqn | r = 2 \beta^2 - \beta^2 
+}}
+{{eqn | r = \beta^2
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By [[Moment Generating Function of Exponential Distribution]], the [[Definition:Moment Generating Function|moment generating function]] $M_X$ of $X$ is given by: 
+:$\map {M_X} t = \dfrac 1 {1 - \beta t}$ 
+From [[Variance as Expectation of Square minus Square of Expectation]]:
+:$\var X = \expect {X^2} - \paren {\expect X}^2$ 
+From [[Moment in terms of Moment Generating Function]], we also have: 
+:$\expect {X^2} = \map {M_X''} 0$
+In [[Expectation of Exponential Distribution/Proof 2|Expectation of Exponential Distribution: Proof 2]], it is shown that: 
+:$\map {M_X'} t = \dfrac \beta {\paren {1 - \beta t}^2}$ 
+We have: 
+{{begin-eqn}}
+{{eqn	| l = \map {M_X''} t
+	| r = \frac \d {\d t} \paren {\frac \beta {\paren {1 - \beta t}^2} }
+}}
+{{eqn	| r = \frac {2 \beta^2} {\paren {1 - \beta t}^3}
+	| c = [[Chain Rule for Derivatives]], [[Derivative of Power]] 
+}}
+{{end-eqn}} 
+Setting $t = 0$ gives: 
+{{begin-eqn}}
+{{eqn	| l = \expect {X^2}
+	| r = \frac {2 \beta^2} {\paren {1 - 0 \beta}^3}
+}}
+{{eqn	| r = 2 \beta^2
+}}
+{{end-eqn}}
+By [[Expectation of Exponential Distribution]], we have: 
+:$\expect X = \beta$
+So: 
+{{begin-eqn}}
+{{eqn	| l = \var X
+	| r = 2 \beta^2 - \beta^2
+}}
+{{eqn	| r = \beta^2
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cardinality of Singleton}
+Tags: Cardinals, Cardinality
+
+\begin{theorem}
+Let $A$ be a [[Definition:Set|set]].
+Then $\left\vert{A}\right\vert = 1$ {{iff}} $\exists a: A = \left\{ {a}\right\}$
+where $\left\vert{A}\right\vert$ denotes the [[Definition:Cardinality|cardinality]] of $A$.
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Assume that
+:$\left\vert{A}\right\vert = 1$
+By definition of [[Definition:Cardinality of Finite Set|cardinality of finite set]]:
+:$A \sim \N_{< 1} = \left\{ {0}\right\}$
+where $\sim$ denotes [[Definition:Set Equivalence|set equivalence]].
+By [[Set Equivalence is Equivalence Relation]]:
+:$\left\{ {0}\right\} \sim A$
+By definition of [[Definition:Set Equivalence|set equivalence]] there exists a [[Definition:Bijection|bijection]]:
+:$f: \left\{ {0}\right\} \to A$
+By definition of [[Definition:Bijection|bijection]]:
+:$f$ is s [[Definition:Surjection|surjection]].
+Thus
+{{begin-eqn}}
+{{eqn | l = A
+      | r = f^\to\left({\left\{ {0}\right\} }\right)
+      | c = {{Defof|Surjection}}
+}}
+{{eqn | r = \left\{ {f\left({0}\right)}\right\}
+      | c = [[Image of Singleton under Mapping]]
+}}
+{{end-eqn}}
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that:
+:$\exists a: A = \left\{ {a}\right\}$
+Define a [[Definition:Mapping|mapping]] $f: A \to \left\{ {0}\right\}$:
+:$F\left({a}\right) = 0$
+It is easy to see by definition that
+:$f$ is an [[Definition:Injection|injection]] and a [[Definition:Surjection|surjection]].
+By definition
+:$f$ is [[Definition:Bijection|bijection]].
+By definition of [[Definition:Set Equivalence|set equivalence]]:
+:$A \sim \left\{ {0}\right\} = \N_{< 1}$
+Thus by definition of [[Definition:Cardinality of Finite Set|cardinality of finite set]]:
+:$\left\vert{A}\right\vert = 1$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Existence of Non-Empty Finite Infima in Meet Semilattice}
+Tags: Join and Meet Semilattices
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Meet Semilattice|meet semilattice]].
+Let $A$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $S$.
+Then $A$ admits a [[Definition:Infimum of Set|infimum]] in $\struct {S, \preceq}$.
+\end{theorem}
+
+\begin{proof}
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]] of the proof of [[Existence of Non-Empty Finite Suprema in Join Semilattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Circle is Bisected by Diameter}
+Tags: Circles, Circle is Bisected by Diameter
+
+\begin{theorem}
+A [[Definition:Circle|circle]] is [[Definition:Bisection|bisected]] by a [[Definition:Diameter of Circle|diameter]].
+\end{theorem}
+
+\begin{proof}
+:[[File:Diameter Bisects Circle.png|400px]]
+Let $AB$ be a [[Definition:Diameter of Circle|diameter]] of a [[Definition:Circle|circle]] whose [[Definition:Center of Circle|center]] is at $O$.
+By definition of [[Definition:Diameter of Circle|diameter]], $AB$ passes through $O$.
+$\angle AOB\cong\angle BOA$ because they are both straight angles.
+Thus, the arcs are congruent by [[Equal Angles in Equal Circles]]:
+:$\stackrel{\frown}{AB}\cong\stackrel{\frown}{BA}$
+Hence, a [[Definition:Circle|circle]] is split into two equal [[Definition:Arc of Circle|arcs]] by a [[Definition:Diameter of Circle|diameter]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $AB$ be a [[Definition:Diameter of Circle|diameter]] of a [[Definition:Circle|circle]] $ADBE$ whose [[Definition:Center of Circle|center]] is at $C$.
+By definition of [[Definition:Diameter of Circle|diameter]], $AB$ passes through $C$.
+{{AimForCont}} that $AB$ does not [[Definition:Bisection|bisect]] $ADBE$, but that $ADBC$ is larger than $AEBC$.
+:[[File:CircleBisectedByDiameter.png|400px]]
+Thus it will be possible to find a [[Definition:Diameter of Circle|diameter]] $DE$ passing through $C$ such that $DC \ne CE$.
+Both $DC$ and $CE$ are [[Definition:Radius of Circle|radii]] of $ADBE$.
+By [[Definition:Euclid's Definitions - Book I/15 - Circle|Euclid's definition of the circle]]:
+:''{{:Definition:Euclid's Definitions - Book I/15 - Circle}}''
+:''{{:Definition:Euclid's Definitions - Book I/16 - Center of Circle}}''
+That is, all [[Definition:Radius of Circle|radii]] of $ADBE$ are equal.
+But $DC \ne CE$.
+From this [[Definition:Contradiction|contradiction]] it follows that $AB$ [[Definition:Bisection|bisects]] the [[Definition:Circle|circle]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Thales' Theorem/Historical Note}
+Tags: Historical Notes
+
+\begin{theorem}
+This result was known by the ancient Babylonians as early as $2000$ BCE.
+The proof of what is now known as [[Thales' Theorem]] that {{AuthorRef|Thales of Miletus}} actually used is unknown.  Whether he was aware of the [[External Angle of Triangle equals Sum of other Internal Angles|Interior Angles Theorem]] and used it in his proof cannot be decided.
+The [[External Angle of Triangle equals Sum of other Internal Angles|Interior Angles Theorem]] is traditionally ascribed to {{AuthorRef|Pythagoras of Samos|Pythagoras}}, who lived near {{AuthorRef|Thales of Miletus|Thales}}, and may have met him.
+Thus it is possible that {{AuthorRef|Pythagoras of Samos|Pythagoras}} learned of the [[External Angle of Triangle equals Sum of other Internal Angles|Interior Angles Theorem]] either directly or indirectly from {{AuthorRef|Thales of Miletus|Thales}} himself.
+\end{theorem}<|endoftext|>
+\section{External Angle of Triangle equals Sum of other Internal Angles}
+Tags: Triangles, Euclid Book I, External Angle of Triangle equals Sum of other Internal Angles
+
+\begin{theorem}
+The [[Definition:External Angle|external angle]] of a [[Definition:Triangle (Geometry)|triangle]] equals the sum of the other two [[Definition:Internal Angle|internal angles]].
+{{:Euclid:Proposition/I/32}}
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-I-32.png|300px]]
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
+Let $BC$ be extended to a point $D$.
+[[Construction of Parallel Line|Construct $CE$]] through the point $C$ [[Definition:Parallel Lines|parallel]] to the [[Definition:Straight Line|straight line]] $AB$.
+We have that $AB \parallel CE$ and $AC$ is a [[Definition:Transversal (Geometry)|transversal]] that cuts them.
+From [[Parallelism implies Equal Alternate Angles]]:
+:$\angle BAC = \angle ACE$
+Similarly, we have that $AB \parallel CE$ and $BD$ is a [[Definition:Transversal (Geometry)|transversal]] that cuts them.
+From [[Parallelism implies Equal Corresponding Angles]]:
+:$\angle ECD = \angle ABC$
+Thus [[Axiom:Euclid's Common Notions|by Euclid's Second Common Notion]]:
+: $\angle ACD = \angle ABC + \angle BAC$
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
+From [[Sum of Angles of Triangle equals Two Right Angles/Proof 2|Sum of Angles of Triangle equals Two Right Angles: Proof 2]]:
+:$\paren 1: \angle ABC + \angle BCA + \angle CAB = 180^\circ$
+Extend $AB$ to $D$.
+By [[Two Angles on Straight Line make Two Right Angles]]:
+:$\paren 1: \angle ABC + \angle CBD = 180^\circ$
+Combining $\paren 1$ and $\paren 2$ and using [[Equality is Transitive]]:
+:$\angle ABC + \angle BCA + \angle CAB = \angle ABC + \angle CBD$
+By using [[Axiom:Euclid's Common Notions|commom notion 3]]:
+:$\angle BCA + \angle CAB = \angle CBD$
+By using [[Equality is Symmetric]]:
+:$\angle CBD = \angle BCA + \angle CAB$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pappus's Centroid Theorem}
+Tags: Solids of Revolution, Surfaces of Revolution
+
+\begin{theorem}
+=== [[Pappus's Centroid Theorem/Surface Area|Pappus's Centroid Theorem for Surface Area]] ===
+{{:Pappus's Centroid Theorem/Surface Area}}
+=== [[Pappus's Centroid Theorem/Volume|Pappus's Centroid Theorem for Volume]] ===
+{{:Pappus's Centroid Theorem/Volume}}
+\end{theorem}<|endoftext|>
+\section{Pappus's Centroid Theorem/Volume}
+Tags: Solids of Revolution
+
+\begin{theorem}
+Let $C$ be a [[Definition:Plane Figure|plane figure]] that lies entirely on one side of a [[Definition:Straight Line|straight line]] $L$.
+Let $S$ be the [[Definition:Solid of Revolution|solid of revolution]] generated by $C$ around $L$.
+Then the [[Definition:Volume|volume]] of $S$ is equal to the [[Definition:Area|area]] of $C$ multiplied by the [[Definition:Distance (Linear Measure)|distance]] travelled by the [[Definition:Centroid|centroid]] of $C$ around $L$ when generating $S$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Pappus of Alexandria}}
+\end{proof}<|endoftext|>
+\section{Pappus's Centroid Theorem/Surface Area}
+Tags: Surfaces of Revolution
+
+\begin{theorem}
+Let $C$ be a [[Definition:Plane Figure|plane figure]] that lies entirely on one side of a [[Definition:Straight Line|straight line]] $L$.
+Let $S$ be the [[Definition:Solid of Revolution|solid of revolution]] generated by $C$ around $L$.
+Then the [[Definition:Surface Area|surface area]] of $S$ is equal to the [[Definition:Perimeter|perimeter length]] of $C$ multiplied by the [[Definition:Distance (Linear Measure)|distance]] travelled by the [[Definition:Centroid|centroid]] of $C$ around $L$ when generating $S$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Pappus of Alexandria}}
+\end{proof}<|endoftext|>
+\section{Pythagoras's Theorem for Parallelograms}
+Tags: Triangles, Parallelograms
+
+\begin{theorem}
+Let $\triangle ABC$ be a [[Definition:Triangle (Geometry)|triangle]].
+Let $ACDE$ and $BCFG$ be [[Definition:Parallelogram|parallelograms]] constructed on the [[Definition:Side of Polygon|sides]] $AC$ and $BC$ of $\triangle ABC$.
+Let $DE$ and $FG$ be [[Definition:Production|produced]] to [[Definition:Intersection (Geometry)|intersect]] at $H$.
+Let $AJ$ and $BI$ be constructed on $A$ and $B$ [[Definition:Parallel Lines|parallel]] to and equal to $HC$.
+Then the [[Definition:Area|area]] of the [[Definition:Parallelogram|parallelogram]] $ABIJ$ equals the sum of the [[Definition:Area|areas]] of the [[Definition:Parallelogram|parallelograms]] $ACDE$ and $BCFG$.
+\end{theorem}
+
+\begin{proof}
+:[[File:PappusPythagorasExtension.png|400px]]
+From [[Parallelograms with Same Base and Same Height have Equal Area]]:
+:$ACDE = ACHR = ATUJ$
+and:
+:$BCFG = BCHS = BIUT$
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping Preserves Finite and Filtered Infima}
+Tags: Order Theory, Join and Meet Semilattices
+
+\begin{theorem}
+Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ be [[Definition:Meet Semilattice|meet semilattices]].
+Let $f: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Let $f$ [[Definition:Mapping Preserves Infimum/Finite|preserve finite infima]] and [[Definition:Mapping Preserves Infimum/Filtered|preserve filtered infima]].
+Then $f$ also [[Definition:Mapping Preserves Infimum/All|preserves all infima]]
+\end{theorem}
+
+\begin{proof}
+Assume that
+:$(1): \quad f$ [[Definition:Mapping Preserves Infimum/Finite|preserves finite infima]]
+and
+:$(2): \quad f$ [[Definition:Mapping Preserves Infimum/Filtered|preserves filtered infima]].
+Let $X$ be a [[Definition:Subset|subset]] of $S_1$.
+Let $X$ admits an [[Definition:Infimum of Set|infimum]] in $\left({S_1, \preceq_1}\right)$
+Define $Z := \left\{ {\inf A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$
+where
+:$\inf A$ denotes the [[Definition:Infimum of Set|infimum]] of $A$ in $\left({S_1, \preceq_1}\right)$
+:${\it Fin}\left({X}\right)$ denotes the set of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $X$
+By [[Existence of Non-Empty Finite Infima in Meet Semilattice]]:
+:for every [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] $A$ of $X$, $A$ admits an [[Definition:Infimum of Set|infimum]] in $\left({S_1, \preceq_1}\right)$
+By [[Infimum of Infima]]:
+:$\inf Z = \inf\bigcup \left({ {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\} }\right) = \inf X$
+:$Z$ admits an [[Definition:Infimum of Set|infimum]] in $\left({S_1, \preceq_1}\right)$
+We will prove that
+:$Z$ is [[Definition:Filtered Subset|filtered]]
+Let $x, y \in Z$.
+By definition of $Z$:
+:$\exists A \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: x = \inf A$
+and
+:$\exists B \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: y = \inf B$
+By [[Finite Union of Finite Sets is Finite]]:
+:$A \cup B$ is [[Definition:Finite Set|finite]]
+:$A \cup B \ne \varnothing$
+By definition of $Z$:
+:$\inf\left({A \cup B}\right) \in Z$
+By [[Infimum of Infima]]:
+:$\left({\inf A}\right) \wedge \left({\inf B}\right) = \inf\left({A \cup B}\right)$
+Thus by definition of [[Definition:Infimum of Set|infimum]]:
+:$\inf\left({A \cup B}\right) \preceq_1 x \land \inf\left({A \cup B}\right) \preceq_1 y$
+Thus by definition
+:$Z$ is [[Definition:Filtered Subset|filtered]]
+By $(2)$ and definitions of [[Definition:Mapping Preserves Infimum/Filtered|mapping preserves filtered infima]] and [[Definition:Mapping Preserves Infimum/Subset|mapping preserves infimum on subset]]:
+:$f^\to\left({Z}\right)$ admits an [[Definition:Infimum of Set|infimum]] in $\left({S_2, \preceq_2}\right)$ and $\inf\left({f^\to\left({Z}\right)}\right) = f\left({\inf Z}\right)$
+We will prove that
+:$X \subseteq Z$
+Let $x \in X$.
+By definition of ${\it Fin}\left({X}\right)$:
+:$\left\{ {x}\right\} \in {\it Fin}\left({X}\right)$ and $\left\{ {x}\right\} \ne \varnothing$
+By definition of $Z$:
+:$\inf \left\{ {x}\right\} \in Z$
+Thus by [[Infimum of Singleton]]:
+:$x \in Z$
+By [[Image of Subset under Relation is Subset of Image/Corollary 2]]:
+:$f^\to\left({X}\right) \subseteq f^\to\left({Z}\right)$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$\inf\left({f^\to\left({Z}\right)}\right)$ is a [[Definition:Lower Bound of Set|lower bound]] for $f^\to\left({Z}\right)$
+By [[Lower Bound is Lower Bound for Subset]]
+:$\inf\left({f^\to\left({Z}\right)}\right)$ is a [[Definition:Lower Bound of Set|lower bound]] for $f^\to\left({X}\right)$
+We will prove that
+:for every [[Definition:Lower Bound of Set|lower bound]] $a$ for $f^\to\left({X}\right)$, $a$ is [[Definition:Lower Bound of Set|lower bound]] for $f^\to\left({Z}\right)$
+Assume that
+:$a$ is [[Definition:Lower Bound of Set|lower bound]] for $f^\to\left({X}\right)$
+Let $x \in f^\to\left({Z}\right)$.
+By definition of [[Definition:Image of Subset under Mapping|image of set]]:
+:$\exists y \in S_1: y \in Z \land f\left({y}\right) = x$
+By definition of $Z$:
+:$\exists A \in {\it Fin}\left({X}\right): y = \inf A \land A \ne \varnothing$
+By $(1)$ and definition of [[Definition:Mapping Preserves Infimum/Finite|mapping preserves finite infima]]:
+:$f$ [[Definition:Mapping Preserves Infimum/Subset|preserves the infimum]] of $A$
+By [[Existence of Non-Empty Finite Infima in Meet Semilattice]]
+:$A$ admits an [[Definition:Infimum of Set|infimum]] in $\left({S_1, \preceq_1}\right)$ and $\inf\left({f^\to\left({A}\right)}\right) = f\left({\inf A}\right) = x$
+By [[Image of Subset under Relation is Subset of Image/Corollary 2]]:
+:$f^\to\left({A}\right) \subseteq f^\to\left({X}\right)$
+By [[Lower Bound is Lower Bound for Subset]]
+:$a$ is [[Definition:Lower Bound of Set|lower bound]] for $f^\to\left({A}\right)$
+Thus by definition of [[Definition:Infimum of Set|infimum]]:
+:$a \preceq_2 x$
+Thus by definition
+:$a$ is [[Definition:Lower Bound of Set|lower bound]] for $f^\to\left({Z}\right)$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:for every [[Definition:Lower Bound of Set|lower bound]] $a$ for $f^\to\left({X}\right)$, $a \preceq_2 f\left({\inf X}\right)$
+Again by definition of [[Definition:Infimum of Set|infimum]]:
+:$f^\to\left({X}\right)$ admits an [[Definition:Infimum of Set|infimum]] in $\left({S_2, \preceq_2}\right)$
+:$\inf\left({f^\to\left({X}\right)}\right) = f\left({\inf X}\right)$
+Thus the result follows by definition of [[Definition:Mapping Preserves Infimum/All|mapping preserves all infima]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Volume of Sphere from Surface Area}
+Tags: Spheres, Volume Formulas
+
+\begin{theorem}
+The [[Definition:Volume|volume]] $V$ of a [[Definition:Sphere (Geometry)|sphere]] of [[Definition:Radius of Sphere|radius]] $r$ is given by:
+:$V = \dfrac {r A} 3$
+where $A$ is the [[Definition:Surface Area|surface area]] of the [[Definition:Sphere (Geometry)|sphere]].
+\end{theorem}
+
+\begin{proof}
+Let the [[Definition:Surface of Solid Figure|surface]] of the [[Definition:Sphere (Geometry)|sphere]] of [[Definition:Radius of Sphere|radius]] $r$ be divided into many small areas.
+If they are made small enough, they can be approximated to [[Definition:Plane Figure|plane figures]].
+Let the [[Definition:Area|areas]] of these [[Definition:Plane Figure|plane figures]] be denoted:
+:$a_1, a_2, a_3, \ldots$
+Let the [[Definition:Sphere (Geometry)|sphere]] of [[Definition:Radius of Sphere|radius]] $r$ be divided into as many [[Definition:Pyramid|pyramids]] whose [[Definition:Apex of Pyramid|apices]] are at the [[Definition:Center of Sphere|center]] and whose [[Definition:Base of Pyramid|bases]] are these areas.
+From [[Volume of Pyramid]], their [[Definition:Volume|volumes]] are:
+:$\dfrac {r a_1} 3, \dfrac {r a_2} 3, \dfrac {r a_3} 3, \ldots$
+The [[Definition:Volume|volume]] $\mathcal V$ of the [[Definition:Sphere (Geometry)|sphere]] is given by the sum of the [[Definition:Volume|volumes]] of each of these [[Definition:Pyramid|pyramids]]:
+{{begin-eqn}}
+{{eqn | l = \mathcal V
+      | r = \dfrac {r a_1} 3 + \dfrac {r a_2} 3 + \dfrac {r a_3} 3 + \cdots
+      | c = 
+}}
+{{eqn | r = \dfrac r 3 \left({a_1 + a_2 + a_3 + \cdots}\right)
+      | c = 
+}}
+{{end-eqn}}
+But $a_1 + a_2 + a_3 + \cdots$ is the [[Definition:Surface Area|surface area]] of the [[Definition:Sphere (Geometry)|sphere]].
+Hence:
+{{begin-eqn}}
+{{eqn | l = \mathcal V
+      | r = \dfrac r 3 \left({a_1 + a_2 + a_3 + \cdots}\right)
+      | c = 
+}}
+{{eqn | r = \dfrac r 3 A
+      | c = 
+}}
+{{end-eqn}}
+It needs to be noted that this proof is intuitive and non-rigorous.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Volume of Solid of Revolution/Historical Note}
+Tags: Historical Notes, Solids of Revolution
+
+\begin{theorem}
+The technique of finding the [[Volume of Solid of Revolution]] by dividing up the [[Definition:Solid of Revolution|solid of revolution]] into many thin disks and approximating them to [[Definition:Cylinder|cylinders]] was devised by {{AuthorRef|Johannes Kepler}} sometime around or after $1612$, reportedly on the occasion of his wedding in $1613$.
+His inspiration was in the problem of finding the [[Definition:Volume|volume]] of wine barrels accurately.
+He published his technique in his $1615$ work {{BookLink|Nova Stereometria Doliorum Vinariorum|Johannes Kepler}} (New Stereometry of Wine Barrels).
+{{AuthorRef|Gottfried Wilhelm von Leibniz}} redefined the problem by applying the techniques of [[Definition:Integral Calculus|integral calculus]] around $1680$.
+\end{theorem}<|endoftext|>
+\section{Factors of Mersenne Number M67}
+Tags: Mersenne Numbers, Classic Problems
+
+\begin{theorem}
+The [[Definition:Mersenne Number|Mersenne number]] $M_{67}$ has the [[Definition:Divisor of Integer|factors]]:
+:$193 \, 707 \, 721$
+:$761 \, 838 \, 257 \, 287$
+\end{theorem}
+
+\begin{proof}
+First we calculate $M_{67}$, which is $2^{67} - 1$:
+:$2^2 = 2 \times 2 = 4$
+{{begin-eqn}}
+{{eqn | l = 2^4
+      | r = 2^2 \times 2^2
+      | c = 
+}}
+{{eqn | r = 4 \times 4
+      | c = 
+}}
+{{eqn | r = 16
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 2^8
+      | r = 2^4 \times 2^4
+      | c = 
+}}
+{{eqn | r = 16 \times 16
+      | c = 
+}}
+{{eqn | r = 256
+      | c = 
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = 2^{16}
+      | r = 2^8 \times 2^8
+      | c = 
+}}
+{{eqn | r = 256 \times 256
+      | c = 
+}}
+{{eqn | r = 65 \, 536
+      | c = 
+}}
+{{end-eqn}}
+as follows:
+<pre>
+   256
+ x 256
+------
+ 1 536
+12 800
+51 200
+------
+65 536
+------
+</pre>
+{{begin-eqn}}
+{{eqn | l = 2^{32}
+      | r = 2^{16} \times 2^{16}
+      | c = 
+}}
+{{eqn | r = 65 \, 536 \times 65 \, 536
+      | c = 
+}}
+{{eqn | r = 4 \, 924 \, 967 \, 296
+      | c = 
+}}
+{{end-eqn}}
+as follows:
+<pre>
+       65 536
+     x 65 536
+      -------
+      393 216
+    1 966 080
+   32 768 000
+  327 680 000
+3 932 160 000
+-------------
+4 924 967 296
+-------------
+</pre>
+{{begin-eqn}}
+{{eqn | l = 2^{64}
+      | r = 2^{32} \times 2^{32}
+      | c = 
+}}
+{{eqn | r = 4 \, 924 \, 967 \, 296 \times 4 \, 924 \, 967 \, 296
+      | c = 
+}}
+{{eqn | r = 18 \, 446 \, 744 \, 073 \, 709 \, 551 \, 616
+      | c = 
+}}
+{{end-eqn}}
+as follows:
+<pre>
+             4 294 967 296
+           x 4 294 967 296
+            --------------
+            25 769 803 776
+           386 547 056 640
+           858 993 459 200
+        30 064 771 072 000
+       257 698 037 760 000
+     3 865 470 566 400 000
+    17 179 869 184 000 000
+   386 547 056 640 000 000
+   858 993 459 200 000 000
+17 179 869 184 000 000 000
+--------------------------
+18 446 744 073 709 551 616
+--------------------------
+ 1 233 444 664 532 221 1
+</pre>
+{{begin-eqn}}
+{{eqn | l = 2^{67}
+      | r = 2^{64} \times 2^3
+      | c = 
+}}
+{{eqn | r = 18 \, 446 \, 744 \, 073 \, 709 \, 551 \, 616 \times 8
+      | c = 
+}}
+{{eqn | r = 147 \, 573 \, 952 \, 589 \, 676 \, 412 \, 928
+      | c = 
+}}
+{{end-eqn}}
+as follows:
+<pre>
+ 18 446 744 073 709 551 616
+x                         8
+---------------------------
+147 573 952 589 676 412 928
+---------------------------
+ 63 355 33  525  74 414 14
+</pre>
+Subtracting $1$ to get $M_{67}$:
+{{begin-eqn}}
+{{eqn | l = 2^{67} - 1
+      | r = 147 \, 573 \, 952 \, 589 \, 676 \, 412 \, 928 - 1
+      | c = 
+}}
+{{eqn | r = 147 \, 573 \, 952 \, 589 \, 676 \, 412 \, 927
+      | c = 
+}}
+{{end-eqn}}
+Now to calculate $193 \, 707 \, 721 \times 761 \, 838 \, 257 \, 287$:
+<pre>
+                193 707 721
+          x 761 838 257 287
+--------------------------
+              1 355 954 047
+             15 496 617 680
+             38 741 544 200
+          1 355 954 047 000
+          9 685 386 050 000
+         38 741 544 200 000
+      1 549 661 768 000 000
+      5 811 231 630 000 000
+    154 966 176 800 000 000
+    193 707 721 000 000 000
+ 11 622 463 260 000 000 000
+135 595 404 700 000 000 000
+---------------------------
+147 573 952 589 676 412 927
+---------------------------
+  1 223 254 435 432 22  1
+</pre>
+Thus:
+{{begin-eqn}}
+{{eqn | l = 2^{67} - 1
+      | r = 147 \, 573 \, 952 \, 589 \, 676 \, 412 \, 927
+      | c = 
+}}
+{{eqn | r = 193 \, 707 \, 721 \times 761 \, 838 \, 257 \, 287
+      | c = 
+}}
+{{end-eqn}}
+as we were to demonstrate.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower Bound is Lower Bound for Subset}
+Tags: Preorder Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be a [[Definition:Preordered Set|preordered set]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $S$ such that
+:$B \subseteq A$
+Let $L$ be an [[Definition:Element|element]] of $S$.
+Let $L$ be a [[Definition:Lower Bound of Set|lower bound]] for $A$.
+Then $L$ is a [[Definition:Lower Bound of Set|lower bound]] for $B$.
+\end{theorem}
+
+\begin{proof}
+Let $L$ be a [[Definition:Lower Bound of Set|lower bound]] for $A$.
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$\forall x \in A: L \preceq x$
+By definition of [[Definition:Subset|subset]]:
+:$\forall x \in B: x \in A$
+Hence:
+:$\forall x \in B: L \preceq x$
+Thus by definition:
+:$L$ is a [[Definition:Lower Bound of Set|lower bound]] for $B$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Bound is Upper Bound for Subset}
+Tags: Preorder Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be a [[Definition:Preordered Set|preordered set]].
+Let $A, B$ be [[Definition:Subset|subsets]] of $S$ such that
+:$B \subseteq A$
+Let $U$ be an [[Definition:Upper Bound of Set|upper bound]] for $A$.
+Then $U$ is an [[Definition:Upper Bound of Set|upper bound]] for $B$.
+\end{theorem}
+
+\begin{proof}
+Assume that:
+: $U$ is [[Definition:Upper Bound of Set|upper bound]] for $A$.
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\forall x \in A: x \preceq U$
+By definition of [[Definition:Subset|subset]]:
+:$\forall x \in B: x \in A$
+Hence:
+:$\forall x \in B: x \preceq U$
+Thus by definition
+: $U$ is sn [[Definition:Upper Bound of Set|upper bound]] for $B$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Largest Rectangle with Given Perimeter is Square}
+Tags: Squares, Rectangles
+
+\begin{theorem}
+Let $\SS$ be the [[Definition:Set|set]] of all [[Definition:Rectangle|rectangles]] with a given [[Definition:Perimeter|perimeter]] $L$.
+The element of $\SS$ with the largest [[Definition:Area|area]] is the [[Definition:Square (Geometry)|square]] with [[Definition:Length (Linear Measure)|length]] of [[Definition:Side of Polygon|side]] $\dfrac L 4$.
+\end{theorem}
+
+\begin{proof}
+Consider an arbitrary element $R$ of $\SS$.
+Let $B$ be half the [[Definition:Perimeter|perimeter]] of $R$.
+Let $x$ be the [[Definition:Length (Linear Measure)|length]] of one [[Definition:Side of Polygon|side]] of $R$.
+Then the [[Definition:Length (Linear Measure)|length]] of an [[Definition:Adjacent Sides|adjacent side]] is $B - x$.
+The [[Definition:Area|area]] $\AA$ of $R$ is then given by:
+:$\AA = x \paren {B - x}$
+Let $\AA$ be expressed in functional notation as:
+:$\map f x = x \paren {B - x}$
+We have:
+{{begin-eqn}}
+{{eqn | l = \map f {x + E} - \map f x
+      | r = \paren {x + E} \paren {B - \paren {x + E} } - x \paren {B - x}
+      | c = 
+}}
+{{eqn | r = E B - 2 E x - E^2
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\map f {x + E} - \map f x} E
+      | r = B - 2 x - E
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \intlimits {\frac {\map f {x + E} - \map f x} E} {E \mathop = 0} {}
+      | r = B - 2 x
+      | c = 
+}}
+{{end-eqn}}
+Thus from [[Derivative at Maximum or Minimum]], when $B - 2 x = 0$, the [[Definition:Area|area]] of $R$ is at a [[Definition:Maximum Value|maximum]] or a [[Definition:Minimum Value|minimum]].
+That is:
+:$x = \dfrac B 2 = \dfrac L 4$
+Geometrical analysis shows that for this [[Definition:Length (Linear Measure)|length]] of [[Definition:Side of Polygon|side]] the [[Definition:Area|area]] is not a [[Definition:Minimum Value|minimum]] because that happens when $x = 0$ or $x = B$.
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Mapping Preserves Finite and Directed Suprema}
+Tags: Order Theory, Join and Meet Semilattices
+
+\begin{theorem}
+Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ be [[Definition:Join Semilattice|join semilattices]].
+Let $f: S_1 \to S_2$ be a [[Definition:Mapping|mapping]].
+Let $f$ [[Definition:Mapping Preserves Supremum/Finite|preserve finite suprema]] and [[Definition:Mapping Preserves Supremum/Directed|preserve directed suprema]].
+Then $f$ also [[Definition:Mapping Preserves Supremum/All|preserves all suprema]]
+\end{theorem}
+
+\begin{proof}
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]] of the proof of [[Mapping Preserves Finite and Filtered Infima]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integral of Power/Historical Note}
+Tags: Historical Notes
+
+\begin{theorem}
+The [[Integral of Power/Conventional Proof|conventional proof]] of [[Integral of Power]] of course holds for all [[Definition:Real Number|real]] $n \ne -1$, not just where $n$ is a [[Definition:Strictly Positive|strictly positive]] [[Definition:Rational Number|rational]].
+However, the real point of this page is [[Integral of Power/Fermat's Proof|Fermat's proof]], which demonstrates how integration was achieved before the full machinery of calculus had been thoroughly constructed.
+{{AuthorRef|Bonaventura Francesco Cavalieri}} had previously made progress with this problem, proving it for integral $1 \le n \le 9$ but the algebra for the proof of each power was more difficult than the previous one, and he found $10$ too much hard work. The clear beauty of {{AuthorRef|Pierre de Fermat|Fermat}}'s approach was that it works for ''all'' $n$, [[Definition:Rational Number|rational]] as well as [[Definition:Integer|integral]].
+\end{theorem}<|endoftext|>
+\section{Integer as Sum of Polygonal Numbers}
+Tags: Polygonal Numbers, Integer as Sum of Polygonal Numbers
+
+\begin{theorem}
+Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
+Then $n$ is:
+:$(1): \quad$ Either [[Definition:Triangular Number|triangular]] or the [[Definition:Integer Addition|sum]] of $2$ or $3$ [[Definition:Triangular Number|triangular numbers]]
+:$(2): \quad$ Either [[Definition:Square Number|square]] or the [[Definition:Integer Addition|sum]] of $2$, $3$ or $4$ [[Definition:Square Number|square numbers]]
+:$(3): \quad$ Either [[Definition:Pentagonal Number|pentagonal]] or the [[Definition:Integer Addition|sum]] of $2$, $3$, $4$ or $5$ [[Definition:Pentagonal Number|pentagonal numbers]]
+:and so on.
+That is:
+:for all $k \ge 3$, $n$ is the [[Definition:Integer Addition|sum]] of no more than $k$ [[Definition:Polygonal Number|polygonal numbers of order $k$]].
+\end{theorem}
+
+\begin{proof}
+First some [[Definition:Lemma|lemmata]]:
+=== [[Integer as Sum of Polygonal Numbers/Lemma 1|Lemma 1]] ===
+{{:Integer as Sum of Polygonal Numbers/Lemma 1}}{{qed|lemma}}
+=== [[Integer as Sum of Polygonal Numbers/Lemma 2|Lemma 2]] ===
+{{:Integer as Sum of Polygonal Numbers/Lemma 2}}{{qed|lemma}}
+=== [[Integer as Sum of Polygonal Numbers/Lemma 3|Lemma 3]] ===
+{{:Integer as Sum of Polygonal Numbers/Lemma 3}}{{qed|lemma}}
+First note that [[Cauchy's Lemma (Number Theory)]] gives us:
+{{:Cauchy's Lemma (Number Theory)}}{{qed|lemma}}
+Now we consider the numbered statements as asserted.
+For $(1)$, we have [[Integer is Sum of Three Triangular Numbers]].
+For $(2)$, we have [[Lagrange's Four Square Theorem]].
+For each $m \ge 3$, we prove that $n$ is the [[Definition:Integer Addition|sum]] of at most $m + 2$ [[Definition:Polygonal Number|polygonal numbers of order $m + 2$]].
+We choose to consider [[Definition:Polygonal Number|polygonal numbers of order $m + 2$]], rather than [[Definition:Polygonal Number|$m$-gonal numbers]], because the former have a neater [[Closed Form for Polygonal Numbers|closed form]]:
+:$\map P {m + 2, k} = \dfrac m 2 \paren {k^2 - k} + k$
+For $n < 116 m$ the result is shown in [[Integer as Sum of Polygonal Numbers/Lemma 1|Lemma 1]].
+It remains to investigate the [[Definition:Integer|integers]] $n \ge 116 m$.
+Define:
+:$I = \openint {\dfrac 2 3 + \sqrt {8 \paren {\dfrac n m} - 8} } {\dfrac 1 2 + \sqrt {6 \paren {\dfrac n m} - 3} }$
+By [[Integer as Sum of Polygonal Numbers/Lemma 2|Lemma 2]]:
+:For $\dfrac n m \ge 116$, the [[Definition:Length of Real Interval|length]] of $I$ is greater than $4$.
+Thus $I$ must contain at least $2$ consecutive [[Definition:Odd Integer|odd integers]].
+Let $b_1, b_2$ be those consecutive [[Definition:Odd Integer|odd integers]].
+Because $b_1 + m - 1 = b_2 + m - 3$:
+:the [[Definition:Set|set]] $\set {b + r: b \in \set {b_1, b_2}, r \in \set {0, 1, \dots, m - 2} }$ contains a [[Definition:Complete Residue System|complete residue system modulo $m$]].
+Hence one can choose some $b, r$ and write $n \equiv b + r \pmod m$.
+Define:
+:$a = 2 \paren {\dfrac {n - b - r} m} + b = \paren {1 - \dfrac 2 m} b + 2 \paren {\dfrac {n - r} m}$
+Since $m \divides \paren {n - b - r}$ and $b$ is [[Definition:Odd Integer|odd]]:
+:$a$ is an [[Definition:Odd Integer|odd integer]].
+Since $b \in I$, from [[Integer as Sum of Polygonal Numbers/Lemma 3|Lemma 3]] we have:
+:$b^2 < 4 a$
+:$3 a < b^2 + 2 b + 4$
+Then by [[Cauchy's Lemma (Number Theory)]], we can write:
+:$a = s^2 + t^2 + u^2 + v^2$
+:$b = s + t + u + v$
+for [[Definition:Non-Negative Integer|nonnegative integers]] $s, t, u, v$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = n
+      | r = \frac m 2 \paren {a - b} + b + r
+      | c = From $a = 2 \paren {\dfrac {n - b - r} m} + b$
+}}
+{{eqn | r = \frac m 2 \paren {s^2 + t^2 + u^2 + v^2 - s - t - u - v} + s + t + u + v + r
+}}
+{{eqn | r = \paren {\frac m 2 \paren {s^2 - s} + s} + \paren {\frac m 2 \paren {t^2 - t} + t} + \paren {\frac m 2 \paren {u^2 - u} + u} + \paren {\frac m 2 \paren {v^2 - v} + v} + r \times 1
+}}
+{{eqn | r = \map P {m + 2, s} + \map P {m + 2, t} + \map P {m + 2, u} + \map P {m + 2, v} + r \map P {m + 2, 1}
+}}
+{{end-eqn}}
+Since $r \le m - 2$, the above expression comprises at most $m + 2$ [[Definition:Polygonal Number|polygonal numbers of order $m + 2$]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infima Preserving Mapping on Filters is Increasing}
+Tags: Order Theory, Increasing Mappings
+
+\begin{theorem}
+Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be [[Definition:Ordered Set|ordered sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+For every [[Definition:Filter in Ordered Set|filter]] $F$ in $\struct {S, \preceq}$, let $f$ [[Definition:Mapping Preserves Infimum/Subset|preserve the infimum]] on $F$.
+Then $f$ is [[Definition:Increasing Mapping|increasing]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$ such that:
+:$x \preceq y$
+By [[Infimum of Singleton]]:
+:$\set x$ and $\set y$ admit [[Definition:Infimum of Set|infima]] in $\struct {S, \preceq}$
+By [[Infimum of Upper Closure of Set]]:
+:$\set x^\succeq$ and $\set y^\succeq$ admit [[Definition:Infimum of Set|infima]] in $\struct {S, \preceq}$
+where $\set x^\succeq$ denotes the [[Definition:Upper Closure of Subset|upper closure]] of $\set x$.
+By [[Upper Closure of Singleton]]
+:$x^\succeq$ and $y^\succeq$ admit [[Definition:Infimum of Set|infima]] in $\struct {S, \preceq}$
+By [[Upper Closure of Element is Filter]]:
+:$x^\succeq$ and $y^\succeq$ are [[Definition:Filter in Ordered Set|filter]] in $\struct {S, \preceq}$
+By assumption and definition of [[Definition:Mapping Preserves Infimum/Subset|mapping preserves the infimum on subset]]:
+:$\map {f^\to} {x^\succeq}$ and $\map {f^\to} {y^\succeq}$ admit [[Definition:Infimum of Set|infima]] in $\struct {T, \precsim}$
+and
+:$\inf \set {\map {f^\to} {x^\succeq} } = \map f {\inf \set {x^\succeq} }$ and $\inf \set {\map {f^\to} {y^\succeq} } = \map f {\inf \set {y^\succeq} }$
+By [[Infimum of Upper Closure of Element]]:
+:$\inf \set {x^\succeq} = x$ and $\inf \set {y^\succeq} = y$
+By [[Upper Closure is Decreasing]]:
+:$y^\succeq \subseteq x^\succeq$
+By [[Image of Subset under Relation is Subset of Image/Corollary 2]]:
+:$\map {f^\to} {y^\succeq} \subseteq \map {f^\to} {x^\succeq}$
+Thus by [[Infimum of Subset]]:
+:$\map f x \precsim \map f y$
+Thus by definition:
+:$f$ is [[Definition:Increasing Mapping|increasing]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infimum of Upper Closure of Set}
+Tags: Upper Closures
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $T \subseteq S$.
+Let $U = T^\succeq$ be the [[Definition:Upper Closure of Subset|upper closure]] of $T$ in $S$.
+Let $s \in S$.
+Then $s$ is the [[Definition:Infimum of Set|infimum]] of $T$ {{iff}} it is the [[Definition:Infimum of Set|infimum]] of $U$.
+\end{theorem}
+
+\begin{proof}
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]] of the proof of [[Supremum of Lower Closure of Set]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Cavalieri's Principle/Extension}
+Tags: Solid Geometry
+
+\begin{theorem}
+Let two [[Definition:Solid Figure|solid figures]] $S_1$ and $S_2$ have equal [[Definition:Height (Linear Measure)|height]].
+Let the [[Definition:Area|areas]] of the [[Definition:Intersection (Geometry)|sections]] made by [[Definition:Plane|planes]] [[Definition:Parallel Planes|parallel]] to their [[Definition:Base of Solid Figure|bases]] and at equal [[Definition:Distance (Linear Measure)|distances]] from the [[Definition:Base of Solid Figure|bases]] always have the same [[Definition:Ratio|ratio]].
+Then the [[Definition:Volume|volumes]] of $S_1$ and $S_2$ are in that same [[Definition:Ratio|ratio]].
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Bonaventura Francesco Cavalieri|cat = Cavalieri}}
+\end{proof}<|endoftext|>
+\section{Cavalieri's Principle}
+Tags: Solid Geometry
+
+\begin{theorem}
+Let two [[Definition:Solid Figure|solid figures]] $S_1$ and $S_2$ have equal [[Definition:Height (Linear Measure)|height]].
+Let [[Definition:Intersection (Geometry)|sections]] made by [[Definition:Plane|planes]] [[Definition:Parallel Planes|parallel]] to their [[Definition:Base of Solid Figure|bases]] and at equal [[Definition:Distance (Linear Measure)|distances]] from the [[Definition:Base of Solid Figure|bases]] always have equal [[Definition:Area|area]].
+Then the [[Definition:Volume|volumes]] of $S_1$ and $S_2$ are equal.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Bonaventura Francesco Cavalieri|cat = Cavalieri}}
+\end{proof}<|endoftext|>
+\section{Volume of Cone}
+Tags: Cones, Volume Formulas
+
+\begin{theorem}
+Let $K$ be a [[Definition:Cone|cone]] whose [[Definition:Base of Cone|base]] is of [[Definition:Area|area]] $A$ and whose [[Definition:Height of Cone|height]] is $h$.
+Then the [[Definition:Volume|volume]] of $K$ is given by:
+:$V_K = \dfrac {A h} 3$
+\end{theorem}
+
+\begin{proof}
+:[[File:Euclid-XII-10.png|300px]]
+Let a [[Definition:Right Circular Cone|cone]] and a [[Definition:Cylinder|cylinder]] have the same [[Definition:Base of Solid Figure|base]] $ABCD$.
+Let them be of equal [[Definition:Height of Solid Figure|height]].
+Suppose the [[Definition:Cylinder|cylinder]] is not three times the [[Definition:Volume|volume]] of the [[Definition:Right Circular Cone|cone]].
+Then the [[Definition:Cylinder|cylinder]] will either be greater than or less than three times the [[Definition:Volume|volume]] of the [[Definition:Right Circular Cone|cone]].
+First suppose that the [[Definition:Cylinder|cylinder]] is greater than three times the [[Definition:Volume|volume]] of the [[Definition:Right Circular Cone|cone]].
+From {{EuclidPropLink|book = IV|prop = 6|title = Inscribing Square in Circle}}:
+:let the [[Definition:Square (Geometry)|square]] $ABCD$ be [[Definition:Polygon Inscribed in Circle|inscribed]] in the [[Definition:Circle|circle]] $ABCD$.
+From [[Square Inscribed in Circle is greater than Half Area of Circle]]:
+:the [[Definition:Square (Geometry)|square]] $ABCD$ is greater than half the [[Definition:Circle|circle]] $ABCD$.
+From the [[Definition:Square (Geometry)|square]] $ABCD$, let a [[Definition:Prism|prism]] be set up whose [[Definition:Height of Prism|height]] equals the [[Definition:Height of Cylinder|height]] of the [[Definition:Cylinder|cylinder]].
+Again using the reasoning of:
+:{{EuclidPropLink|book = XII|prop = 2|title = Areas of Circles are as Squares on Diameters}}
+in conjunction with 
+:{{EuclidPropLink|book = XI|prop = 32|title = Parallelepipeds of Same Height have Volume Proportional to Bases}}
+this [[Definition:Prism|prism]] is greater than half the [[Definition:Cylinder|cylinder]].
+Let the [[Definition:Arc of Circle|arcs]] $AB, BC, CD, DA$ be [[Definition:Bisection|bisected]] at $E, F, G, H$.
+Let $AE, EB, BF, FC, CG, GD, DH, HA$ be joined.
+From {{EuclidPropLink|book = XII|prop = 2|title = Areas of Circles are as Squares on Diameters}}:
+:each of $\triangle AEB, \triangle BFC, \triangle CGD, \triangle DHA$ is greater than half the [[Definition:Segment of Circle|segment]] of the [[Definition:Circle|circle]] $ABCD$ around it.
+On each of $\triangle AEB, \triangle BFC, \triangle CGD, \triangle DHA$, let [[Definition:Prism|prisms]] be set up with the same [[Definition:Height of Solid Figure|height]] as the [[Definition:Cylinder|cylinder]].
+Again from the reasoning in {{EuclidPropLink|book = XII|prop = 2|title = Areas of Circles are as Squares on Diameters}}:
+:each of these [[Definition:Prism|prisms]] is greater than half the [[Definition:Segment of Circle|segment]] of the [[Definition:Cylinder|cylinder]] around it.
+The operation:
+: [[Definition:Bisection|bisecting]] the [[Definition:Arc of Circle|arcs]] remaining, joining the points of bisection with [[Definition:Straight Line|straight lines]] and setting up [[Definition:Prism|prisms]] on the resulting [[Definition:Triangle (Geometry)|triangles]] the same [[Definition:Height of Solid Figure|height]] as the [[Definition:Cylinder|cylinder]]
+can be repeated indefinitely.
+From {{EuclidPropLink|book = X|prop = 1|title = Existence of Fraction of Number Smaller than Given}}:
+: eventually we will leave some [[Definition:Segment of Circle|segments]] of the [[Definition:Cylinder|cylinder]] which will be less than the amount by which the [[Definition:Cylinder|cylinder]] is greater than triple the [[Definition:Right Circular Cone|cone]].
+Let such [[Definition:Segment of Circle|segments]] be left, and let them be $AE, EB, BF, FC, CG, GD, DH, HA$.
+The remainder, the [[Definition:Prism|prism]] whose [[Definition:Base of Prism|base]] is the [[Definition:Polygon|polygon]] $AEBFCGDH$ and whose [[Definition:Height of Prism|height]] equals the [[Definition:Height of Cylinder|height]] of the [[Definition:Cylinder|cylinder]], is then greater than three times the [[Definition:Right Circular Cone|cone]].
+But from {{EuclidPropLink|book = XII|prop = 7|qual=Porism|title = Prism on Triangular Base divided into Three Equal Tetrahedra}}:
+:the [[Definition:Prism|prism]] whose [[Definition:Base of Prism|base]] is the [[Definition:Polygon|polygon]] $AEBFCGDH$ and whose [[Definition:Height of Prism|height]] equals the [[Definition:Height of Cylinder|height]] of the [[Definition:Cylinder|cylinder]] is three times the [[Definition:Pyramid|pyramid]] whose [[Definition:Base of Pyramid|base]] is $AEBFCGDH$ and whose [[Definition:Apex of Pyramid|apex]] is the same as the [[Definition:Apex of Cone|apex]] of the [[Definition:Right Circular Cone|cone]].
+Therefore the [[Definition:Pyramid|pyramid]] whose [[Definition:Base of Pyramid|base]] is $AEBFCGDH$ and whose [[Definition:Apex of Pyramid|apex]] is the same as the [[Definition:Apex of Cone|apex]] of the [[Definition:Right Circular Cone|cone]] is greater than the [[Definition:Right Circular Cone|cone]] which has the [[Definition:Circle|circle]] $ABCD$ as base.
+But it is also less, as it is enclosed by it.
+From this impossibility it follows that the [[Definition:Cylinder|cylinder]] cannot be greater than three times the [[Definition:Volume|volume]] of the [[Definition:Right Circular Cone|cone]].
+Now suppose that the [[Definition:Cylinder|cylinder]] is less than three times the [[Definition:Volume|volume]] of the [[Definition:Right Circular Cone|cone]].
+That is, the [[Definition:Right Circular Cone|cone]] is greater than a third part of the [[Definition:Cylinder|cylinder]].
+From {{EuclidPropLink|book = IV|prop = 6|title = Inscribing Square in Circle}}:
+:let the [[Definition:Square (Geometry)|square]] $ABCD$ be [[Definition:Polygon Inscribed in Circle|inscribed]] in the [[Definition:Circle|circle]] $ABCD$.
+From [[Square Inscribed in Circle is greater than Half Area of Circle]]:
+:the [[Definition:Square (Geometry)|square]] $ABCD$ is greater than half the [[Definition:Circle|circle]] $ABCD$.
+From the [[Definition:Square (Geometry)|square]] $ABCD$, let a [[Definition:Pyramid|pyramid]] be set up whose [[Definition:Apex of Pyramid|apex]] is the same as the [[Definition:Apex of Cone|apex]] of the [[Definition:Right Circular Cone|cone]].
+Again using the reasoning of:
+:{{EuclidPropLink|book = XII|prop = 2|title = Areas of Circles are as Squares on Diameters}}
+in conjunction with the reasoning of:
+:{{EuclidPropLink|book = XI|prop = 32|title = Parallelepipeds of Same Height have Volume Proportional to Bases}}
+this [[Definition:Pyramid|pyramid]] is greater than half the [[Definition:Right Circular Cone|cone]].
+Let the [[Definition:Arc of Circle|arcs]] $AB, BC, CD, DA$ be [[Definition:Bisection|bisected]] at $E, F, G, H$.
+Let $AE, EB, BF, FC, CG, GD, DH, HA$ be joined.
+From {{EuclidPropLink|book = XII|prop = 2|title = Areas of Circles are as Squares on Diameters}}:
+:each of $\triangle AEB, \triangle BFC, \triangle CGD, \triangle DHA$ is greater than half the [[Definition:Segment of Circle|segment]] of the [[Definition:Circle|circle]] $ABCD$ around it.
+On each of $\triangle AEB, \triangle BFC, \triangle CGD, \triangle DHA$, let [[Definition:Pyramid|pyramids]] be set up whose [[Definition:Apex of Pyramid|apex]] is the same as the [[Definition:Apex of Cone|apex]] of the [[Definition:Right Circular Cone|cone]].
+Again from the reasoning in {{EuclidPropLink|book = XII|prop = 2|title = Areas of Circles are as Squares on Diameters}}:
+:each of these [[Definition:Pyramid|pyramids]] is greater than half the [[Definition:Segment of Circle|segment]] of the [[Definition:Right Circular Cone|cone]] around it.
+The operation:
+: [[Definition:Bisection|bisecting]] the [[Definition:Arc of Circle|arcs]] remaining, joining the points of bisection with [[Definition:Straight Line|straight lines]] and setting up [[Definition:Pyramid|pyramids]] on the resulting [[Definition:Triangle (Geometry)|triangles]] whose [[Definition:Apex of Pyramid|apex]] is the same as the [[Definition:Apex of Cone|apex]] of the [[Definition:Right Circular Cone|cone]]
+can be repeated indefinitely.
+From {{EuclidPropLink|book = X|prop = 1|title = Existence of Fraction of Number Smaller than Given}}:
+: eventually we will leave some [[Definition:Segment of Circle|segments]] of the [[Definition:Right Circular Cone|cone]] which will be less than the amount by which the [[Definition:Right Circular Cone|cone]] is greater than one third the [[Definition:Cylinder|cylinder]].
+Let such [[Definition:Segment of Circle|segments]] be left, and let them be $AE, EB, BF, FC, CG, GD, DH, HA$.
+The remainder, the [[Definition:Pyramid|pyramid]] whose [[Definition:Base of Pyramid|base]] is the [[Definition:Polygon|polygon]] $AEBFCGDH$ and whose [[Definition:Apex of Pyramid|apex]] is the same as the [[Definition:Apex of Cone|apex]] of the [[Definition:Right Circular Cone|cone]], is then greater than one third the [[Definition:Cylinder|cylinder]].
+But from {{EuclidPropLink|book = XII|prop = 7|qual=Porism|title = Prism on Triangular Base divided into Three Equal Tetrahedra}}:
+:the [[Definition:Pyramid|pyramid]] whose [[Definition:Base of Pyramid|base]] is the [[Definition:Polygon|polygon]] $AEBFCGDH$ and whose [[Definition:Apex of Pyramid|apex]] is the same as the [[Definition:Apex of Cone|apex]] of the [[Definition:Right Circular Cone|cone]] is one third the [[Definition:Prism|prism]] whose [[Definition:Base of Prism|base]] is $AEBFCGDH$ and whose [[Definition:Height of Prism|height]] equals the [[Definition:Height of Cylinder|height]] of the [[Definition:Cylinder|cylinder]].
+Therefore the [[Definition:Prism|prism]] whose [[Definition:Base of Prism|base]] is $AEBFCGDH$ and whose [[Definition:Height of Prism|height]] equals the [[Definition:Height of Cylinder|height]] of the [[Definition:Cylinder|cylinder]] is greater than the [[Definition:Cylinder|cylinder]] which has the [[Definition:Circle|circle]] $ABCD$ as base.
+But it is also less, as it is enclosed by it.
+From this impossibility it follows that the [[Definition:Cylinder|cylinder]] cannot be less than three times the [[Definition:Volume|volume]] of the [[Definition:Right Circular Cone|cone]].
+The result follows.
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Let the [[Definition:Cone|cone]] be of [[Definition:Height of Cone|height]] $h$.
+Let the [[Definition:Area|area]] of the [[Definition:Base of Cone|base]] of the [[Definition:Cone|cone]] be $A$.
+From [[Volume of Cylinder]], the [[Definition:Volume|volume]] of a [[Definition:Cylinder|cylinder]] of [[Definition:Base of Cylinder|base]] $A$ and [[Definition:Height of Cylinder|height]] $h$ is $A h$.
+Let the [[Definition:Cone|cone]] be divided by [[Definition:Plane|planes]] [[Definition:Parallel Planes|parallel]] to its [[Definition:Base of Cone|base]] each positioned some small [[Definition:Distance (Linear Measure)|distance]] $d$ apart.
+Let $d$ be sufficiently small that they can be approximated to [[Definition:Cylinder|cylinders]] in shape.
+Let there be $n$ of these small [[Definition:Cylinder|cylinders]] in total.
+Let the [[Definition:Volume|volumes]] of each of these small [[Definition:Cylinder|cylinders]] be:
+: $v_1, v_2, v_3, \ldots, v_n$
+starting from the [[Definition:Base of Cone|base]] of the [[Definition:Cone|cone]] and working up.
+We have that:
+:$v_k = d a_k$
+where $a_n$ is the 
+Let the [[Definition:Area|areas]] of the [[Definition:Base of Cylinder|bases]] of each of these small [[Definition:Cylinder|cylinders]] be:
+: $a_1, a_2, a_3, \ldots, a_n$
+starting from the [[Definition:Base of Cone|base]] of the [[Definition:Cone|cone]] and working up.
+{{improve|clumsy}}
+\end{proof}
+
+\begin{proof}
+Let $V_K$ be the [[Definition:Volume|volume]] of $K$.
+Let $V_C$ be the [[Definition:Volume|volume]] of a [[Definition:Cylinder|cylinder]] of [[Definition:Base of Cylinder|base]] $A$ and of [[Definition:Height of Cylinder|height]] $h$.
+From [[Volume of Cylinder]]:
+:$V_C = A h$
+From [[Volume of Cone is Third of Cylinder on Same Base and of Same Height]]:
+{{begin-eqn}}
+{{eqn | l = V_K
+      | r = \dfrac {V_C} 3
+      | c = 
+}}
+{{eqn | r = \dfrac {A h} 3
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Cones]]
+[[Category:Volume Formulas]]
+l2jfjl03yymwq4tnmi2q8qaskbw69s9
+\end{proof}<|endoftext|>
+\section{Volume of Cylinder}
+Tags: Cylinders, Volume Formulas
+
+\begin{theorem}
+The [[Definition:Volume|volume]] $V_C$ of a [[Definition:Cylinder|cylinder]] whose [[Definition:Base of Cylinder|bases]] are [[Definition:Circle|circles]] of [[Definition:Radius of Circle|radius]] $r$ and whose [[Definition:Height of Cylinder|height]] is $h$ is given by the formula:
+:$V_C = \pi r^2 h$
+\end{theorem}
+
+\begin{proof}
+:[[File:VolumeOfCylinder.png|600px]]
+Consider a [[Definition:Cylinder|cylinder]] $C$ whose [[Definition:Base of Cylinder|base]] is a [[Definition:Circle|circle]] of [[Definition:Radius of Circle|radius]] $r$ and whose [[Definition:Height of Cylinder|height]] is $h$.
+Consider a [[Definition:Cuboid|cuboid]] $K$ whose [[Definition:Height of Cylinder|height]] is $h$ and whose [[Definition:Base of Solid Figure|base]] has the same [[Definition:Area|area]] as the [[Definition:Base of Cylinder|base]] of $C$.
+Let the [[Definition:Area|area]] of those [[Definition:Base of Solid Figure|bases]] be $A$.
+Let the [[Definition:Cylinder|cylinder]] $C$ be positioned with its [[Definition:Base of Cylinder|base]] in the same [[Definition:Plane|plane]] as the [[Definition:Base of Solid Figure|base]] of $K$.
+By [[Cavalieri's Principle]] $C$ and $K$ have the same [[Definition:Volume|volume]].
+The [[Definition:Base of Cylinder|bases]] of $C$ are [[Definition:Circle|circles]] of [[Definition:Radius of Circle|radius]] $r$.
+From [[Area of Circle]], the [[Definition:Area|area]] of each [[Definition:Base of Cylinder|base]] therefore gives:
+:$A = \pi r^2$
+From [[Volume of Cuboid]], $K$ has [[Definition:Volume|volume]] given by:
+:$V_K = A h = \pi r^2 h$
+Hence the result.
+{{qed}}
+[[Category:Cylinders]]
+[[Category:Volume Formulas]]
+oy3b69d3hb9fkft0wxnc17jp4r6u9ea
+\end{proof}<|endoftext|>
+\section{Slope of Tangent to Cycloid}
+Tags: Cycloids, Slope of Tangent to Cycloid
+
+\begin{theorem}
+Let $C$ be a [[Definition:Cycloid|cycloid]] generated by the equations:
+:$x = a \paren {\theta - \sin \theta}$
+:$y = a \paren {1 - \cos \theta}$
+The [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] to $C$ at the point $\tuple {x, y}$ is given by:
+:$\dfrac {\d y} {\d x} = \cot \dfrac \theta 2$
+\end{theorem}
+
+\begin{proof}
+By definition of [[Derivative of Curve at Point]], the [[Definition:Tangent to Curve|tangent]] to $C$ at the point $\tuple {x, y}$ is the [[Definition:Derivative of Real Function at Point|derivative]] of its equation at that point.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \frac {\d x} {\d \theta}
+      | r = a \paren {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | l = \frac {\d y} {\d \theta}
+      | r = a \sin \theta
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\d y} {\d x}
+      | r = \frac {\d y} {\d \theta} / \frac {\d x} {\d \theta}
+      | c = 
+}}
+{{eqn | r = \frac {a \sin \theta} {a \paren {1 - \cos \theta} }
+      | c = 
+}}
+{{eqn | r = \frac {\sin \theta} {1 - \cos \theta}
+      | c = 
+}}
+{{eqn | r = \frac 1 {\map \tan {\theta / 2} }
+      | c = [[Half Angle Formulas/Tangent/Corollary 2|Half Angle Formula for Tangent: Corollary 2]]
+}}
+{{eqn | r = \cot \frac \theta 2
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Consider a [[Definition:Polygon|polygon]] $ABCD$ being rolled along a [[Definition:Straight Line|straight line]] in the same way as the [[Definition:Generating Circle of Cycloid|generating circle]] of $C$.
+Let $A', B', C', D'$ be the [[Definition:Point|points]] around which the $ABCD$ rotates while rolling.
+:[[File:TangentToCycloid-construction.png|600px]]
+The point $A$ traces out in succession several [[Definition:Arc of Circle|arcs of circles]] with [[Definition:Center of Circle|centers]] $B', C', D'$.
+The [[Definition:Tangent to Curve|tangent]] to each of these [[Definition:Arc of Circle|arcs]] is [[Definition:Perpendicular|perpendicular]] to the [[Definition:Straight Line|line]] joining the [[Definition:Point|point]] of [[Definition:Tangent to Curve|tangency]] to the corresponding [[Definition:Point|point]] of rotation.
+Consider a [[Definition:Circle|circle]] as being a [[Definition:Polygon|polygon]] with an [[Definition:Infinity|infinite]] number of [[Definition:Side of Polygon|sides]].
+It follows that in the limit the [[Definition:Tangent to Curve|tangent]] to $C$ at any given [[Definition:Point|point]] is [[Definition:Perpendicular|perpendicular]] to the [[Definition:Straight Line|line]] joining the [[Definition:Point|point]] of [[Definition:Tangent to Curve|tangency]] to the corresponding [[Definition:Point|point]] where the [[Definition:Generating Circle of Cycloid|generating circle]] touches the [[Definition:Straight Line|straight line]] along which the [[Definition:Generating Circle of Cycloid|generating circle]] rolls.
+:[[File:TangentToCycloid.png|600px]]
+Let $O$ be the [[Definition:Center of Circle|center]] of the [[Definition:Generating Circle of Cycloid|generating circle]].
+Let $P$ be the [[Definition:Point|point]] of [[Definition:Tangent to Curve|tangency]].
+Let $Q$ be the [[Definition:Point|point]] where the [[Definition:Generating Circle of Cycloid|generating circle]] touches the [[Definition:Straight Line|straight line]] along which the [[Definition:Generating Circle of Cycloid|generating circle]] rolls.
+Let $S$ be the [[Definition:Point|point]] where the [[Definition:Tangent to Curve|tangent]] to $C$ meets the [[Definition:Straight Line|straight line]] along which the [[Definition:Generating Circle of Cycloid|generating circle]] rolls.
+As $OP$ and $OQ$ are both [[Definition:Radius of Circle|radii]] of the [[Definition:Generating Circle of Cycloid|generating circle]]:
+:$OP = OQ = a$
+and so $\triangle POQ$ is [[Definition:Isosceles Triangle|isosceles]].
+Let $R$ be dropped [[Definition:Perpendicular|perpendicular]] to $PQ$ at $R$.
+As $\triangle POQ$ is [[Definition:Isosceles Triangle|isosceles]], $\triangle OPR = \triangle OQR$ and so $PR = RQ$.
+Thus $\angle POR = \angle QOR = \dfrac \theta 2$.
+As $OR \perp PR$ and $PR \perp PS$, we have that $\triangle OPR$ is [[Definition:Similar Triangles|similar]] to $\triangle QPS$.
+Thus $\angle PQS = \angle POR$ and so $\angle PSQ$ is $\dfrac \pi 2 - \angle POR = \dfrac \pi 2 - \dfrac \theta 2$.
+By definition, $\tan \angle PSQ$ is the [[Definition:Slope|slope]] of the [[Definition:Tangent to Curve|tangent]] $\dfrac {\d y} {\d x}$ to $C$.
+Thus:
+{{begin-eqn}}
+{{eqn | l = \dfrac {\d y} {\d x}
+      | r = \tan \angle PSQ
+      | c = 
+}}
+{{eqn | r = \map \tan {\dfrac \pi 2 - \dfrac \theta 2}
+      | c = 
+}}
+{{eqn | r = \cot \frac \theta 2
+      | c = [[Tangent of Complement equals Cotangent]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Closure of Singleton}
+Tags: Upper Closures
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $s$ be an [[Definition:Element|element]] of $S$.
+Then:
+:$\left\{ {s}\right\}^\succeq = s^\succeq$
+where:
+:$\left\{ {s}\right\}^\succeq$ denotes the [[Definition:Upper Closure/Set|upper closure]] of $\left\{ {s}\right\}$
+:$s^\succeq$ denotes the [[Definition:Upper Closure/Element|upper closure]] of $s$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left\{ {s}\right\}^\succeq
+      | r = \bigcup \left\{ {t^\succeq: t \in \left\{ {s}\right\} }\right\}
+      | c = definition of [[Definition:Upper Closure/Set|upper closure of set]]
+}}
+{{eqn | r = \bigcup \left\{ {s^\succeq}\right\}
+      | c = definition of [[Definition:Singleton|singleton]]
+}}
+{{eqn | r = s^\succeq
+      | c = [[Union of Singleton]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower Closure of Singleton}
+Tags: Lower Closures
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $s$ be an [[Definition:Element|element]] of $S$.
+Then:
+: $\left\{ {s}\right\}^\preceq = s^\preceq$
+where:
+:$\left\{ {s}\right\}^\preceq$ denotes the [[Definition:Lower Closure/Set|lower closure]] of $\left\{ {s}\right\}$
+:$s^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $s$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \left\{ {s}\right\}^\preceq
+      | r = \bigcup \left\{ {t^\preceq: t \in \left\{ {s}\right\} }\right\}
+      | c = Definition of [[Definition:Lower Closure/Set|lower closure of set]]
+}}
+{{eqn | r = \bigcup \left\{ {s^\preceq}\right\}
+      | c = Definition of [[Definition:Singleton|Singleton]]
+}}
+{{eqn | r = s^\preceq
+      | c = [[Union of Singleton]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Closure of Element is Filter}
+Tags: Upper Closures
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $s$ be an [[Definition:Element|element]] of $S$.
+Then:
+:$s^\succeq$ is a [[Definition:Filter in Ordered Set|filter]] in $\struct {S, \preceq}$
+where $s^\succeq$ denotes the [[Definition:Upper Closure/Element|upper closure]] of $s$.
+\end{theorem}
+
+\begin{proof}
+By [[Singleton is Directed and Filtered Subset]]
+:$\set s$ is a [[Definition:Filtered Subset|filtered subset]] of $S$
+By [[Filtered iff Upper Closure Filtered]]:
+:$\set s^\succeq$ is a [[Definition:Filtered Subset|filtered subset]] of $S$
+By [[Upper Closure is Upper Set]]:
+:$\set s^\succeq$ is a [[Definition:Upper Set|upper set]] in $S$
+By [[Upper Closure of Singleton]]
+:$\set s^\succeq = s^\succeq$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$s \preceq s$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$s \in s^\succeq$
+Thus by definition:
+:$s^\succeq$ is [[Definition:Non-Empty Set|non-empty]], [[Definition:Filtered Subset|filtered]] and [[Definition:Upper Set|upper]].
+Thus by definition:
+:$s^\succeq$ is a [[Definition:Filter in Ordered Set|filter]] in $\struct {S, \preceq}$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Logarithmic Spiral}
+Tags: Spirals
+
+\begin{theorem}
+Consider a [[Definition:Logarithmic Spiral|logarithmic spiral]] $S$ given by the equation:
+:$r = a e^{b \theta}$
+Construct a [[Definition:Tangent to Curve|tangent]] to $S$ at the point $Q = \left({a, 0}\right)$.
+Let the [[Definition:Tangent to Curve|tangent]] cross the [[Definition:Y-Axis|$y$-axis]] at $P$.
+Then the [[Definition:Length (Linear Measure)|length]] of $PQ$ equals the total [[Definition:Arc Length|length]] of $S$ from $P$ inwards to the [[Definition:Origin|origin]].
+\end{theorem}
+
+\begin{proof}
+&nbsp;
+:[[File:LogarithmicSpiralLength.png|400px]]
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Logarithmic Spiral is Equiangular}
+Tags: Spirals
+
+\begin{theorem}
+The [[Definition:Logarithmic Spiral|logarithmic spiral]] is '''equiangular''', in the following sense:
+Let $P = \left\langle{r, \theta}\right\rangle$ be a [[Definition:Point|point]] on a [[Definition:Logarithmic Spiral|logarithmic spiral]] $S$ expressed in [[Definition:Polar Coordinates|polar coordinates]] as:
+:$r = a e^{b \theta}$
+Then the [[Definition:Angle|angle]] $\psi$ that the [[Definition:Tangent to Curve|tangent]] makes to the [[Definition:Radius Vector|radius vector]] of $S$ is [[Definition:Constant|constant]].
+\end{theorem}
+
+\begin{proof}
+Consider the [[Definition:Logarithmic Spiral|logarithmic spiral]] $S$ expressed as:
+:$r = a e^{b \theta}$
+:[[File:LogarithmicSpiralAngle.png|400px]]
+Let $\psi$ be the [[Definition:Angle|angle]] between the [[Definition:Tangent to Curve|tangent]] to $S$ and the [[Definition:Radius Vector|radius vector]].
+The  [[Definition:Derivative|derivative]] of $r$ {{WRT}} $\theta$ is:
+:$\dfrac {\mathrm d r} {\mathrm d \theta} = a b e^{b \theta} = b r$
+and thus:
+{{begin-eqn}}
+{{eqn | l = \tan \psi
+      | r = \frac r {\frac {\mathrm d r} {\mathrm d \theta} }
+      | c = 
+}}
+{{eqn | r = \frac r {b r}
+      | c = 
+}}
+{{eqn | r = \frac 1 b
+      | c = 
+}}
+{{eqn | ll= \leadsto
+      | l = \cot \psi
+      | r = b
+      | c = 
+}}
+{{end-eqn}}
+Thus for a given [[Definition:Logarithmic Spiral|logarithmic spiral]], $\psi$ is [[Definition:Constant|constant]] and equal to $\operatorname{arccot} b$.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Length of Logarithmic Spiral/Historical Note}
+Tags: Historical Notes
+
+\begin{theorem}
+The [[Definition:Arc Length|length]] of a [[Definition:Logarithmic Spiral|logarithmic spiral]] was first found by {{AuthorRef|Evangelista Torricelli}} in $1645$.
+This was the first time anybody had found the [[Definition:Arc Length|length]] of a non-[[Definition:Straight Line|straight-line]] [[Definition:Curve|curve]] for anything other than a [[Definition:Circle|circle]].
+Before this had been done, few people could accept that this was possible to do.
+For example, {{AuthorRef|René Descartes}} had stated in his {{BookLink|La Géométrie|René Descartes}} in $1637$:
+:''Geometry should not include lines that are like strings, in that they are sometimes straight and sometimes curved, since the ratios between straight and curved lines are not known, and I believe cannot be discovered by human minds.''
+{{AuthorRef|Galileo}}'s response was:
+:'' Who is so blind as not to see that, if there are two equal straight lines, one of which is then bent into a curve, that curve will be equal to the straight line?''
+\end{theorem}<|endoftext|>
+\section{Infimum of Upper Closure of Element}
+Tags: Upper Closures
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $s$ be an [[Definition:Element|element]] of $S$.
+Then:
+:$\map \inf {s^\succeq} = s$
+where $s^\succeq$ denotes the [[Definition:Upper Closure/Element|upper closure]] of $s$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \inf {s^\succeq}
+      | r = \map \inf {\set s^\succeq}
+      | c = [[Upper Closure of Singleton]]
+}}
+{{eqn | r = \map \inf {\set s}
+      | c = [[Infimum of Upper Closure of Set]]
+}}
+{{eqn | r = s
+      | c = [[Infimum of Singleton]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Volume of Gabriel's Horn}
+Tags: Gabriel's Horn
+
+\begin{theorem}
+Consider [[Definition:Gabriel's Horn|Gabriel's horn]], the [[Definition:Solid of Revolution|solid of revolution]] formed by rotating about the [[Definition:X-Axis|$x$-axis]] the [[Definition:Curve|curve]]:
+:$y = \dfrac 1 x$
+Consider the [[Definition:Volume|volume]] $V$ of the space enclosed by the [[Definition:Plane|planes]] $x = 1$, $x = a$ and the portion of [[Definition:Gabriel's Horn|Gabriel's horn]] where $1 \le x \le a$.
+Then:
+:$V = \pi \left({1 - \dfrac 1 a}\right)$
+\end{theorem}
+
+\begin{proof}
+From [[Volume of Solid of Revolution]]:
+{{begin-eqn}}
+{{eqn | l = V
+      | r = \pi \int_1^a \frac 1 {x^2} \, \mathrm d x
+      | c = 
+}}
+{{eqn | r = \pi \left[{-\dfrac 1 x}\right]_1^a
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \pi \left[{\dfrac 1 x}\right]_a^1
+      | c = 
+}}
+{{eqn | r = \pi \left({1 - \dfrac 1 a}\right)
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Pascal's Theorem}
+Tags: Conic Sections, Projective Geometry
+
+\begin{theorem}
+Let $ABCDEF$ be a [[Definition:Hexagon|hexagon]] whose $6$ [[Definition:Vertex of Polygon|vertices]] lie on a [[Definition:Conic Section|conic section]] and whose [[Definition:Opposite Sides|opposite sides]] are not [[Definition:Parallel Lines|parallel]].
+Then the [[Definition:Point|points]] of [[Definition:Intersection (Geometry)|intersection]] of the [[Definition:Opposite Sides|opposite sides]], when [[Definition:Production|produced]] as necessary, all lie on a [[Definition:Straight Line|straight line]].
+\end{theorem}
+
+\begin{proof}
+:[[File:PascalsTheorem.png|400px]]
+:[[File:PascalsTheorem2.png|500px]]
+{{ProofWanted}}
+\end{proof}<|endoftext|>
+\section{Lower Closure of Element is Ideal}
+Tags: Lower Closures
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ ne an [[Definition:Ordered Set|ordered set]].
+Let $s$ be an [[Definition:Element|element]] of $S$.
+Then $s^\preceq$ is [[Definition:Ideal in Ordered Set|ideal]] in $\left({S, \preceq}\right)$
+where $s^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $s$.
+\end{theorem}
+
+\begin{proof}
+By [[Singleton is Directed and Filtered Subset]]
+:$\left\{ {s}\right\}$ is a [[Definition:Directed Subset|directed subset]] of $S$
+By [[Directed iff Lower Closure Directed]]:
+:$\left\{ {s}\right\}^\preceq$ is a [[Definition:directed Subset|directed subset]] of $S$
+By [[Lower Closure is Lower Set]]:
+:$\left\{ {s}\right\}^\preceq$ is a [[Definition:Lower Set|lower set]] in $S$
+By [[Lower Closure of Singleton]]
+:$\left\{ {s}\right\}^\preceq = s^\preceq$
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$s \preceq s$
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$s \in s^\preceq$
+Thus by definition:
+:$s^\succeq$ is [[Definition:Non-Empty Set|non-empty]] [[Definition:Directed Subset|directed]] and [[Definition:Lower Set|lower]].
+Thus by definition:
+:$s^\preceq$ is a [[Definition:Ideal in Ordered Set|ideal]] in $\left({S, \preceq}\right)$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Supremum of Lower Closure of Element}
+Tags: Lower Closures
+
+\begin{theorem}
+Let $\left({S. \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Let $s$ be an [[Definition:Element|element]] of $S$.
+Then:
+:$\sup \left({s^\preceq}\right) = s$
+where $s^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $s$.
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \sup \left({s^\preceq}\right)
+      | r = \sup \left({\left\{ {s}\right\}^\preceq}\right)
+      | c = [[Lower Closure of Singleton]]
+}}
+{{eqn | r = \sup \left({\left\{ {s}\right\} }\right)
+      | c = [[Supremum of Lower Closure of Set]]
+}}
+{{eqn | r = s
+      | c = [[Supremum of Singleton]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Upper Closure is Decreasing}
+Tags: Upper Closures
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $x, y$ be [[Definition:Element|elements]] of $S$ such that
+:$x \preceq y$
+then $y^\succeq \subseteq x^\succeq$
+where $y^\succeq$ denotes the [[Definition:Upper Closure of Element|upper closure]] of $y$.
+\end{theorem}
+
+\begin{proof}
+Let $z \in y^\succeq$.
+By definition of [[Definition:Upper Closure of Element|upper closure of element]]:
+:$y \preceq z$
+By definition of [[Definition:Ordering|ordering]], $\preceq$ is [[Definition:Transitive Relation|transitive]].
+From $x \preceq y$ and $y \preceq z$:
+:$x \preceq z$
+Thus again by definition of [[Definition:Upper Closure of Element|upper closure of element]]:
+:$z \in x^\succeq$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Lower Closure is Increasing}
+Tags: Lower Closures
+
+\begin{theorem}
+Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
+Let $x, y$ be [[Definition:Element|elements]] of $S$ such that
+:$x \preceq y$
+then $x^\preceq \subseteq y^\preceq$
+where $y^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $y$.
+\end{theorem}
+
+\begin{proof}
+Let $z \in x^\preceq$.
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$z \preceq x$
+By definition of [[Definition:Transitivity|transitivity]]:
+:$z \preceq y$
+Thus again by definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$z \in y^\preceq$
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Suprema Preserving Mapping on Ideals is Increasing}
+Tags: Order Theory, Increasing Mappings
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+For every [[Definition:Ideal in Ordered Set|ideal]] $I$ in $\left({S, \preceq}\right)$, let $f$ [[Definition:Mapping Preserves Supremum/Subset|preserve the supremum]] on $I$.
+Then $f$ is [[Definition:Increasing Mapping|increasing]].
+\end{theorem}
+
+\begin{proof}
+Let $x, y \in S$ such that:
+:$x \preceq y$
+By [[Supremum of Singleton]]:
+:$\left\{ {x}\right\}$ and $\left\{ {y}\right\}$ admit [[Definition:Supremum of Set|suprema]] in $\left({S, \preceq}\right)$
+By [[Supremum of Lower Closure of Set]]:
+:$\left\{ {x}\right\}^\preceq$ and $\left\{ {y}\right\}^\preceq$ admit [[Definition:Supremum of Set|suprema]] in $\left({S, \preceq}\right)$
+where $\left\{ {x}\right\}^\preceq$ denotes the [[Definition:Lower Closure/Set|lower closure]] of $\left\{ {x}\right\}$
+By [[Lower Closure of Singleton]]:
+:$x^\preceq$ and $y^\preceq$ admit [[Definition:Supremum of Set|suprema]] in $\left({S, \preceq}\right)$
+By [[Lower Closure of Element is Ideal]]:
+:$x^\preceq$ and $y^\preceq$ are [[Definition:Ideal in Ordered Set|ideals]] in $\left({S, \preceq}\right)$
+By assumption and definition of [[Definition:Mapping Preserves Supremum/Subset|mapping preserves the supremum on subset]]:
+:$f^\to \left({x^\preceq}\right)$ and $f^\to \left({y^\preceq}\right)$ admit [[Definition:Supremum of Set|suprema]] in $\left({T, \precsim}\right)$
+and
+:$\sup \left({f^\to \left({x^\preceq}\right)}\right) = f \left({\sup \left({x^\preceq}\right)}\right)$ and $\sup \left({f^\to \left({y^\preceq}\right)}\right) = f \left({\sup \left({y^\preceq}\right)}\right)$
+By [[Supremum of Lower Closure of Element]]:
+:$\sup \left({x^\preceq}\right) = x$ and $\sup \left({y^\preceq}\right) = y$
+By [[Lower Closure is Increasing]]:
+:$x^\preceq \subseteq y^\preceq$
+By [[Image of Subset under Relation is Subset of Image/Corollary 2]]:
+:$f^\to \left({x^\preceq}\right) \subseteq f^\to \left({y^\preceq}\right)$
+Thus by [[Supremum of Subset]]:
+:$f\left({x}\right) \precsim f\left({y}\right)$
+Thus by definition:
+:$f$ is [[Definition:Increasing Mapping|increasing]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Imaginary Unit to Power of Itself}
+Tags: Complex Numbers
+
+\begin{theorem}
+:$i^i = e^{-\pi / 2}$
+where $i$ is the [[Definition:Imaginary Unit|imaginary unit]].
+Its [[Definition:Decimal Expansion|decimal expansion]] is:
+:$0 \cdotp 20787 \, 95763 \, 50761 \, 90854 \, 6955 \ldots$
+{{OEIS|A049006}}
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = i^i
+      | r = \left({e^{i \pi / 2} }\right)^i
+      | c = [[Euler's Formula]]
+}}
+{{eqn | r = e^{\pi i^2 / 2}
+      | c = 
+}}
+{{eqn | r = e^{-\pi / 2}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Imaginary Unit to Power of Itself/Complete}
+Tags: Complex Numbers
+
+\begin{theorem}
+:$i^i = \set {\exp \paren {\dfrac {4 k + 3} 2 \pi}: k \in \Z}$
+where $i$ is the [[Definition:Imaginary Unit|imaginary unit]].
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = i^i
+      | r = \exp \paren {i \ln \paren i}
+      | c = {{Defof|Complex Power}}
+}}
+{{eqn | r = \exp \paren {i \paren {\ln 1 + i \paren {\dfrac \pi 2 + 2 k \pi} } }
+      | c = {{Defof|Complex Natural Logarithm}}: for all $k \in \Z$
+}}
+{{eqn | r = \exp \paren {i \paren {i \paren {\dfrac \pi 2 + 2 k \pi} } } 
+      | c = [[Logarithm of 1 is 0]]
+}}
+{{eqn | r = \exp \paren {i^2 \dfrac \pi 2 + 2 k \pi i^2}
+      | c = 
+}}
+{{eqn | r = \exp \paren {-\dfrac \pi 2 - 2 k \pi}
+      | c = 
+}}
+{{eqn | r = \exp \paren {\dfrac {3 \pi} 2 + 2 k \pi}
+      | c = as $k$ ranges over all [[Definition:Integer|integers]]
+}}
+{{eqn | r = \exp \paren {\dfrac {4 k + 3} 2 \pi}
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Riemann Zeta Function of 4}
+Tags: Formulas for Pi, Riemann Zeta Function at Even Integers, Riemann Zeta Function of 4, Fourth Powers
+
+\begin{theorem}
+The [[Definition:Riemann Zeta Function|Riemann zeta function]] of $4$ is given by:
+{{begin-eqn}}
+{{eqn | l = \map \zeta 4
+      | r = \dfrac 1 {1^4} +  \dfrac 1 {2^4} +  \dfrac 1 {3^4} +  \dfrac 1 {4^4} + \cdots
+      | c = 
+}}
+{{eqn | r = \dfrac {\pi^4} {90}
+      | c = 
+}}
+{{eqn | o = \approx
+      | r = 1 \cdotp 08232 \, 3 \ldots
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+By [[Fourier Series of Fourth Power of x]], for $x \in \closedint {-\pi} \pi$:
+:$\displaystyle x^4 = \frac {\pi^4} 5 + \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \, \map \cos {n \pi} \, \map \cos {n x}$
+Setting $x = \pi$: 
+{{begin-eqn}}
+{{eqn | l = \pi^4 
+      | r = \frac {\pi^4} 5 + \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2 - 48} {n^4} \, \map {\cos^2} {n \pi}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {4 \pi^4} 5 
+      | r = \sum_{n \mathop = 1}^\infty \frac {8 n^2 \pi^2} {n^4} - \sum_{n \mathop = 1}^\infty \frac {48} {n^4}
+      | c = [[Cosine of Multiple of Pi]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\pi^4} 5
+      | r = 2 \pi^2 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4}
+}}
+{{eqn | r = \frac {\pi^4} 3 - 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4}
+      | c = [[Basel Problem]]
+}}
+{{eqn | ll= \leadsto
+      | l = 12 \sum_{n \mathop = 1}^\infty \frac 1 {n^4} 
+      | r = \frac {\pi^4} 3 - \frac {\pi^4} 5
+      | c = rearranging
+}}
+{{eqn | r = \frac {2 \pi^4} {15}
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^4}
+      | r = \frac {\pi^4} {90}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+By [[Fourier Series of x squared]], for $x \in \left[{- \pi \,.\,.\, \pi}\right]$:
+:$\displaystyle x^2 = \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\left({-1}\right)^n \frac 4 {n^2} \cos n x}\right)$
+Hence:
+{{begin-eqn}}
+{{eqn | l = \frac 1 \pi \int_{-\pi}^\pi x^4 \, \mathrm d x
+      | r = \frac 1 2 \left({\frac {2 \pi^2} 3}\right)^2 + \sum_{n \mathop = 1}^\infty \left({\frac {4 \left({-1}\right)^n} {n^2} }\right)^2
+      | c = [[Parseval's Theorem]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac 2 \pi \int_0^\pi x^4 \, \mathrm d x 
+      | r = \frac {2 \pi^4} 9 + \sum_{n \mathop = 1}^\infty \frac {16} {n^4} 
+      | c = [[Definite Integral of Even Function]] 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {2 \pi^4} 5 
+      | r = \frac {2 \pi^4} 9 + \sum_{n \mathop = 1}^\infty \frac {16} {n^4} 
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {8 \pi^4} {45} 
+      | r = \sum_{n \mathop = 1}^\infty \frac {16} {n^4}
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^4} 
+      | r = \frac {\pi^4} {90}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{ProofWanted|[[Riemann Zeta Function as a Multiple Integral|$\displaystyle \int_0^1\int_0^1\int_0^1\int_0^1 \frac 1 {1 - x_1x_2x_3x_4} \mathrm dx_1 \mathrm dx_2 \mathrm dx_3 \mathrm dx_4$]]. Use Beukers-Calabi-Kolk sub.}}
+\end{proof}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \map \zeta 4
+      | r = \paren {-1}^3 \dfrac {B_4 2^3 \pi^4} {4!}
+      | c = [[Riemann Zeta Function at Even Integers]]
+}}
+{{eqn | r = \paren {-1}^3 \paren {-\dfrac 1 {30} } \dfrac {2^3 \pi^4} {4!}
+      | c = {{Defof|Sequence of Bernoulli Numbers}}
+}}
+{{eqn | r = \paren {\dfrac 1 {30} } \paren {\dfrac 8 {24} } \pi^4
+      | c = {{Defof|Factorial}}
+}}
+{{eqn | r = \dfrac {\pi^4} {90}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+From the [[Basel Problem]] we have:
+:$\displaystyle \map \zeta 2 = \sum_{n \mathop = 1}^{\infty} {\frac 1 {n^2}} = \frac {\pi^2} 6$
+Therefore:
+{{begin-eqn}}
+{{eqn | l = \map \zeta 2
+      | r = \paren {\frac 1 {1^2 } + \frac 1 {2^2 } + \frac 1 {3^2 } + \frac 1 {4^2 } + \cdots }
+      | c = 
+}}
+{{eqn | r = \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }
+      | c = 
+}}
+{{end-eqn}}
+Taking the square of $\map \zeta 2$, we have:
+{{begin-eqn}}
+{{eqn | l = \displaystyle \paren {\map \zeta 2}^2
+      | r = \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots } \paren {1 + \frac 1 4 + \frac 1 9 + \frac 1 {16 } + \cdots }
+      | c =
+}}
+{{eqn | r = \paren {\sum_{i \mathop = 1}^{\infty} {\frac 1 {i^2 } } } \paren {\sum_{j \mathop = 1}^{\infty} {\frac 1 {j^2 } } }
+      | c =
+}}
+{{eqn | r = \sum_{i \mathop = 1}^{\infty} \sum_{j \mathop = 1}^{\infty} {\frac 1 {i^2 } } {\frac 1 {j^2 } }
+      | c = [[Product of Absolutely Convergent Series]]
+}}
+{{eqn | r = \sum_{i \mathop = 1}^{\infty} {\frac 1 {i^4 } } + \sum_{i \mathop = 2}^{\infty} \sum_{j \mathop = 1}^{i-1} {\frac 1 {i^2 } } {\frac 1 {j^2 } } + \sum_{j \mathop = 2}^{\infty} \sum_{i \mathop = 1}^{j - 1} {\frac 1 {i^2 } } {\frac 1 {j^2 } }
+      | c = $\paren {i = j } + \paren {j \lt i } + \paren {j \gt i }$
+}}
+{{eqn | r = \map \zeta 4  + 2  \dfrac {\pi^4} {5!}
+      | c = The sums $\paren {j < i }$ and $\paren {j > i }$ are [[Definition:Symmetric Relation|symmetric]] and each equal to the [[Definition:Coefficient of Polynomial/Polynomial Form|coefficient]] of the 5th power term in the [[Definition:Sine/Complex Function|sin(x)]] expansion See [[Basel_Problem/Proof_2]]
+}}
+{{end-eqn}}
+:$\begin{array}{r|cccccccccc}
+\displaystyle \paren {\map \zeta 2}^2 & \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {16}}  & \cdots \\
+\hline
+\paren {\dfrac {1} {1}} & \paren {\dfrac {1} {1}}^2 & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {1}} \paren {\dfrac {1} {16}} & \cdots \\
+\paren {\dfrac {1} {4}}  & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {4}}^2 & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {4}} \paren {\dfrac {1} {16}} & \cdots \\
+\paren {\dfrac {1} {9}}  & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {9}}^2 & \paren {\dfrac {1} {9}} \paren {\dfrac {1} {16}} & \cdots \\
+\paren {\dfrac {1} {16}}  & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {1}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {4}} & \paren {\dfrac {1} {16}} \paren {\dfrac {1} {9}} & \paren {\dfrac {1} {16}}^2 & \cdots \\
+\cdots  & \cdots & \cdots & \cdots & \cdots & \cdots \\
+\end{array}$
+{{begin-eqn}}
+{{eqn | l = \map \zeta 4
+      | r = \paren {\map \zeta 2}^2  - 2  \dfrac {\pi^4} {5!}
+      | c = Rearranging
+}}
+{{eqn | r = \dfrac {\pi^4} {36} - \dfrac {\pi^4} {60}
+      | c = simplifying
+}}
+{{eqn | r = \dfrac {\pi^4} {90}
+      | c = simplifying
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Riemann Zeta Function of 6}
+Tags: Riemann Zeta Function at Even Integers, Formulas for Pi, Sixth Powers, Riemann Zeta Function of 6
+
+\begin{theorem}
+The [[Definition:Riemann Zeta Function|Riemann zeta function]] of $6$ is given by:
+{{begin-eqn}}
+{{eqn | l = \map \zeta 6
+      | r = \dfrac 1 {1^6} +  \dfrac 1 {2^6} +  \dfrac 1 {3^6} +  \dfrac 1 {4^6} + \cdots
+      | c = 
+}}
+{{eqn | r = \dfrac {\pi^6} {945}
+      | c = 
+}}
+{{eqn | o = \approx
+      | r = 1 \cdotp 01734 \, 3 \ldots
+      | c = 
+}}
+{{end-eqn}}
+\end{theorem}
+
+\begin{proof}
+By [[Fourier Series/Sixth Power of x over Minus Pi to Pi|Fourier Series: $x^6$ over $-\pi$ to $\pi$]], for $x \in \closedint {-\pi} \pi$:
+:$\displaystyle x^6 = \frac {\pi^6} 7 + \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 +1440} {n^6} \, \map \cos {n \pi} \, \map \cos {n x}$
+Setting $x = \pi$: 
+{{begin-eqn}}
+{{eqn | l = \pi^6 
+      | r = \frac {\pi^6} 7 + \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 +1440} {n^6} \, \map {\cos^2} {n \pi}
+}}
+{{eqn | ll= \leadsto
+      | l = \frac { 6 \pi^6} 7 
+      | r = \sum_{n \mathop = 1}^\infty \frac {12 \pi^4} {n^2} - \sum_{n \mathop = 1}^\infty \frac {240 \pi^2} {n^4} + \sum_{n \mathop = 1}^\infty \frac {1440} {n^6}
+      | c = [[Cosine of Multiple of Pi]]
+}}
+{{eqn | ll= \leadsto
+      | l = \frac {\pi^6} 7
+      | r = 2 \pi^4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2} - 40 \pi^2 \sum_{n \mathop = 1}^\infty \frac 1 {n^4} + 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6}
+}}
+{{eqn | r = - \frac {\pi^6} 9 + 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6}
+      | c = [[Basel Problem]] and [[Riemann Zeta Function of 4]]
+}}
+{{eqn | ll= \leadsto
+      | l = 240 \sum_{n \mathop = 1}^\infty \frac 1 {n^6} 
+      | r = \frac {\pi^6} 9 + \frac {\pi^6} 7
+      | c = rearranging
+}}
+{{eqn | r = \frac {16 \pi^4} {63}
+}}
+{{eqn | ll= \leadsto
+      | l = \sum_{n \mathop = 1}^\infty \frac 1 {n^6}
+      | r = \frac {\pi^6} {945}
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}
+
+\begin{proof}
+{{tidy|use the equation template consistently}}
+{{begin-eqn}}
+{{eqn | l = \sin x
+      | r = x \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }
+      | c = [[Euler Formula for Sine Function]]
+}}
+{{eqn | l = \displaystyle \dfrac {\sin x} x
+      | r = \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} } = \paren {1 - \dfrac {x^2} {1^2 \pi^2} } \paren {1 - \dfrac {x^2} {2^2 \pi^2} } \paren {1 - \dfrac {x^2} {3^2 \pi^2} } \cdots
+}}
+{{end-eqn}}
+{{begin-eqn}}
+{{eqn | l = \sin x
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n + 1} } {\paren {2 n + 1}!} = x - \dfrac {x^3} {3!} + \dfrac {x^5} {5!} - \dfrac {x^7} {7!} + \cdots
+      | c = [[Power Series Expansion for Sine Function]]
+}}
+{{eqn | l = \displaystyle \dfrac {\sin x} x
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \dfrac {x^{2 n} } {\paren {2 n + 1}!} = 1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots
+}}
+{{end-eqn}}
+Equating the product with the sum:
+{{begin-eqn}}
+{{eqn | l = \displaystyle \dfrac {\sin x} x
+      | r = \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }
+      | rr = = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {x^{2 n} } {\paren {2 n + 1}!}
+}}
+{{eqn | r = \paren {1 - \dfrac {x^2} {1 \pi^2} } x \paren {1 - \dfrac {x^2} {4 \pi^2} } x \paren {1 - \dfrac {x^2} {9 \pi^2} } x \cdots
+      | rr = = 1 - \dfrac {x^2} {3!} + \dfrac {x^4} {5!} - \dfrac {x^6} {7!} + \cdots
+}}
+{{end-eqn}}
+Each squared term in the product selected once:
+{{begin-eqn}}
+{{eqn | l = \displaystyle -\dfrac {x^2} {3!}
+      | r = -\dfrac {x^2} {\pi^2} \paren {1 + \dfrac 1 4 + \dfrac 1 9 + \dfrac 1 {16} + \cdots}
+      | c = Term raised to the second power used to calculate the [[Riemann Zeta Function at Even Integers/Examples/2|Riemann Zeta Function of $2$]]
+}}
+{{end-eqn}}
+Each unique combination of '''two''' squared terms in the product selected:
+:$\dfrac {x^4} {5!} = \dfrac {x^4} { \pi^4}  
+\paren {\paren 1 \paren {\dfrac 1 4} + \paren 1 \paren {\dfrac 1 9} + \paren 1 \paren {\dfrac 1 {16}} + \cdots +
+\paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren {\dfrac 1 4} \paren {\dfrac 1 {16}} + \cdots + 
+\paren {\dfrac 1 9} \paren {\dfrac 1 {16} } }$ Term raised to the fourth power used to calculate [[Riemann Zeta Function at Even Integers/Examples/4|Riemann Zeta Function of $4$]]
+Each unique combination of '''three''' squared terms in the product selected:
+:$-\dfrac {x^6} {7!} = -\dfrac {x^6} {\pi^6}  
+\paren {\paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 9} + \paren 1 \paren {\dfrac 1 4} \paren {\dfrac 1 {16} } + \cdots +
+\paren 1 \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots + \paren {\dfrac 1 4} \paren {\dfrac 1 9} \paren {\dfrac 1 {16} } + \cdots}$ Term raised to the sixth used power to calculate [[Riemann Zeta Function at Even Integers/Examples/6|Riemann Zeta Function of $6$]]
+{{begin-eqn}}
+{{eqn | l = \map \zeta 2
+      | r = \dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots
+      | rr = = \dfrac {\pi^2} 6
+      | c = [[Basel Problem]]
+}}
+{{eqn | l = \paren {\map \zeta 2}^3
+      | r = \paren {\dfrac 1 {1^2} + \dfrac 1 {2^2} + \dfrac 1 {3^2} + \dfrac 1 {4^2} + \cdots}^3
+      | rr = = \paren {\dfrac {\pi^2} 6}^3 = \dfrac {\pi^6} {216}
+}}
+{{end-eqn}}
+When we take the cube of a sum, we have:
+{{begin-eqn}}
+{{eqn | l = \paren {A + B + C + \cdots}^3
+      | r = \paren {A^3 + B^3 + C^3 + \cdots} + 3 \paren {A^2 B + A B^2 + A^2 C + A C^2 + B^2 C + B C^2 + \cdots} + 6 \paren {A B C + \cdots}
+      | c = 
+}}
+{{end-eqn}}
+Let $A = \dfrac 1 {1^2}, B = \dfrac 1 {2^2}, C = \dfrac 1 {3^2}, \cdots $
+Then the {{LHS}} becomes:
+:$\paren {\paren {\dfrac 1 {1^2} } + \paren {\dfrac 1 {2^2} } + \paren {\dfrac 1 {3^2} } + \cdots}^3 = \paren {\map \zeta 2}^3$
+and the first term on the {{RHS}} becomes:
+:$\paren {\paren {\dfrac 1 {1^2} }^3 + \paren {\dfrac 1 {2^2} }^3 + \paren {\dfrac 1 {3^2} }^3 + \cdots} = \map \zeta 6$
+To make sense of the remaining two terms on the {{RHS}}, we need:
+:$\paren {AB + AC + BC + \cdots} = \dfrac {\pi^4} {5!} $
+:$\paren {A + B + C + \cdots} = \dfrac {\pi^2} {3!} $
+{{begin-eqn}}
+{{eqn | l = \paren {AB + AC + BC + \cdots}  \paren {A + B + C + \cdots}
+      | r = \paren {A^2B + AB^2 + A^2C + AC^2 + B^2C + BC^2 + \cdots} + 3\paren {ABC + \cdots}
+      | c = 
+}}
+{{eqn | l = 3 \paren {AB + AC + BC + \cdots} \paren {A + B + C + \cdots}
+      | r = 3 \paren {A^2B + AB^2 + A^2C + AC^2 + B^2C + BC^2 + \cdots} + 9 \paren {ABC + \cdots}
+      | c = We have 3 too many of the 'ABC' type (only need 6) - need to subtract 3 of these
+}}
+{{end-eqn}}
+Finally, we have:
+{{begin-eqn}}
+{{eqn | l = \paren {\map \zeta 2}^3
+      | r = \map \zeta 6  + 3 \dfrac {\pi^2} {3!} \dfrac {\pi^4} {5!} - 3  \dfrac {\pi^6} {7!}
+      | c =
+}}
+{{eqn | ll= \leadsto
+      | l = \map \zeta 6
+      | r = \paren {\map \zeta 2}^3  - 3  \dfrac {\pi^2} {3!} \dfrac {\pi^4} {5!} + 3  \dfrac {\pi^6} {7!}
+      | c = rearranging
+}}
+{{eqn | r = \dfrac {\pi^6} {216} - 21 \dfrac {\pi^6} {7!} + 3 \dfrac {\pi^6} {7!}
+      | c = simplifying
+}}
+{{eqn | r = 70 \dfrac {\pi^6} {3 x 7!} - 54 \dfrac {\pi^6} {3 x 7!}
+      | c = simplifying
+}}
+{{eqn | r = 16 \dfrac {\pi^6} {3 x 7!}
+      | c = simplifying
+}}
+{{eqn | r = \dfrac {\pi^6} {945}
+      | c =
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Infima Preserving Mapping on Filters Preserves Filtered Infima}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be [[Definition:Ordered Set|ordered sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+For every [[Definition:Filter in Ordered Set|filter]] $F$ in $\struct {S, \preceq}$, let $f$ [[Definition:Mapping Preserves Infimum/Subset|preserve the infimum]] on $F$.
+Then $f$ [[Definition:Mapping Preserves Infimum/Filtered|preserves filtered infima]].
+\end{theorem}
+
+\begin{proof}
+Let $F$ be a [[Definition:Filtered Subset|filtered]] [[Definition:Subset|subset]] of $S$ such that:
+:$F$ admits an [[Definition:Infimum of Set|infimum]] in $\struct {S, \preceq}$.
+By [[Filtered iff Upper Closure Filtered]]:
+:$F^\succeq$ is [[Definition:Filtered Subset|filtered]]
+where $F^\succeq$ denotes the [[Definition:Upper Closure/Set|upper closure]] of $F$.
+By [[Upper Closure is Upper Set]]:
+:$F^\succeq$ is [[Definition:Upper Set|upper]].
+Because [[Definition:Filtered Subset|filtered]] is [[Definition:Non-Empty Set|non-empty]], by definition:
+:$F^\succeq$ is [[Definition:Filter in Ordered Set|filter]] in $\struct {S, \preceq}$.
+By [[Infimum of Upper Closure of Set]]:
+:$F^\succeq$ admits an [[Definition:Infimum of Set|infimum]] in $\struct {S, \preceq}$
+and
+:$\map \inf {F^\succeq} = \inf F$
+By assumption and [[Definition:Mapping Preserves Infimum/Subset|mapping preserves the infimum on subset]]:
+:$\map {f^\to} {F^\succeq}$ admits an [[Definition:Infimum of Set|infimum]] in $\struct {T, \precsim}$
+and
+:$\map \inf {\map {f^\to} {F^\succeq} } = \map f {\map \inf {F^\succeq} }$
+By [[Upper Closure is Closure Operator]]:
+:$F \subseteq F^\succeq$
+By [[Image of Subset under Relation is Subset of Image/Corollary 2]]:
+:$\map {f^\to} F \subseteq \map {f^\to} {F^\succeq}$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$\map f {\inf F}$ is [[Definition:Lower Bound of Set|lower bound]] for $\map {f^\to} {F^\succeq}$
+By [[Lower Bound is Lower Bound for Subset]]:
+:$\map f {\inf F}$ is a [[Definition:Lower Bound of Set|lower bound]] for $\map {f^\to} F$
+We will prove that:
+:for every [[Definition:Element|element]] $x$ of $T$:
+::if $x$ is [[Definition:Lower Bound of Set|lower bound]] for $\map {f^\to} F$, then $x \precsim \map f {\inf F}$
+Let $x \in T$ such that:
+:$x$ is a [[Definition:Lower Bound of Set|lower bound]] for $\map {f^\to} \F$
+We will prove as a sublemma that:
+:$x$ is a [[Definition:Lower Bound of Set|lower bound]] for $\map {f^\to} {F^\succeq}$
+Let $y \in \map {f^\to} {F^\succeq}$.
+By definition of [[Definition:Image of Subset under Mapping|image of set under mapping]]:
+:$\exists a \in S: a \in F^\succeq \land y = \map f a$
+By definition of [[Definition:Upper Closure/Set|upper closure of set]]:
+:$\exists b \in F: b \preceq a$
+By [[Infima Preserving Mapping on Filters is Increasing]]:
+:$f$ is [[Definition:Increasing Mapping|increasing]].
+By definition of [[Definition:Increasing Mapping|increasing]]:
+:$\map f b \precsim \map f a$
+By definition of [[Definition:Image of Subset under Mapping|image of set under mapping]]:
+:$\map f b \in \map {f^\to} F$
+By definition of [[Definition:Lower Bound of Set|lower bound of set]]:
+:$x \precsim \map f b$
+Thus by definition of [[Definition:Transitivity|transitivity]]:
+:$x \precsim y$
+Thus by definition:
+:$x$ is a [[Definition:Lower Bound of Set|lower bound]] for $\map {f^\to} {F^\succeq}$
+This ends the proof of the sublemma.
+Thus by definition of [[Definition:Infimum of Set|infimum]]:
+:$x \precsim \map f {\inf F}$
+Thus again by definition of [[Definition:Infimum of Set|infimum]]:
+:$\map {f^\to} F$ admits an [[Definition:Infimum of Set|infimum]] in $\struct {T, \precsim}$
+and
+:$\map \inf {\map {f^\to} F} = \map f {\inf F}$
+Thus result follows by definition of [[Definition:Mapping Preserves Infimum/Filtered|mapping preserves filtered infima]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Approximate Formula for Number of Partitions}
+Tags: Partition Theory
+
+\begin{theorem}
+The [[Definition:Partition Function|number of partitions]] $\map p n$ of a [[Definition:Strictly Positive Integer|(strictly) positive integer]] $n$ is given by the approximation:
+:$\map p n \approx \dfrac {e^{\pi \sqrt {2 n / 3} } } {4 n \sqrt 3}$
+\end{theorem}<|endoftext|>
+\section{Number of Partitions as Coefficient of Power Series}
+Tags: Partition Theory
+
+\begin{theorem}
+The [[Definition:Partition Function|number of partitions]] $\map p n$ of a [[Definition:Strictly Positive Integer|(strictly) positive integer]] $n$ is equal to the [[Definition:Coefficient|coefficient]] of $x^n$ when the [[Definition:Expression|expression]]:
+:$\map f n = \dfrac 1 {\paren {1 - x} \paren {1 - x^2} \paren {1 - x^3} \cdots}$
+is expanded into a [[Definition:Power Series|power series]].
+That is:
+:$\map f n = 1 + \map p 1 x + \map p 2 x^2 + \map p 3 x^3 + \cdots$
+or:
+:$\displaystyle \sum_{n \mathop \in \Z_{\ge 0} } \map p n q^n = \prod_{j \mathop \in \Z_{>0} } \dfrac 1 {1 - q^j}$
+where $\map p 0 := 1$.
+\end{theorem}<|endoftext|>
+\section{Suprema Preserving Mapping on Ideals Preserves Directed Suprema}
+Tags: Order Theory
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be [[Definition:Ordered Set|ordered sets]].
+Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
+Let every [[Definition:Filter in Ordered Set|filter]] $F$ in $\left({S, \preceq}\right)$, $f$ [[Definition:Mapping Preserves Infimum/Subset|preserve the infimum]] on $F$.
+Then $f$ [[Definition:Mapping Preserves Supremum/Directed|preserves directed suprema]].
+\end{theorem}
+
+\begin{proof}
+This follows by [[Definition:Mutatis Mutandis|mutatis mutandis]] of the proof of [[Infima Preserving Mapping on Filters Preserves Filtered Infima]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Euler's Equations of Motion for Rotation of Rigid Body}
+Tags: Definitions: Classical Mechanics
+
+\begin{theorem}
+Let a [[Definition:Rigid|rigid]] [[Definition:Body|body]] $B$ [[Definition:Space Rotation|rotate]] about an [[Definition:Axis|axis]] $A$ which is fixed in relation to $B$ and [[Definition:Parallel Lines|parallel]] to the [[Definition:Principal Axis of Inertia|principal axis of inertia]] of $B$.
+Then the [[Definition:Space Rotation|rotation]] of $B$ about $A$ is described by:
+:$\mathbf I \cdot \dot {\boldsymbol \omega} + \boldsymbol \omega \times \left({\mathbf I \cdot \boldsymbol\omega}\right) = \mathbf M$
+where:
+:$\mathbf M$ is the applied [[Definition:Torque|torque]] applied to $B$ about $A$
+:$\mathbf I$ is the [[Definition:Moment of Inertia|moment of inertia]] of $B$ with respect to $A$
+:$\boldsymbol \omega$ is the [[Definition:Angular Velocity|angular velocity]] about $A$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Leonhard Paul Euler|cat = Euler}}
+\end{proof}<|endoftext|>
+\section{Euler's Hydrodynamical Equation for Flow of Ideal Incompressible Fluid}
+Tags: Hydrodynamics
+
+\begin{theorem}
+:$\begin{cases} \dfrac {D \mathbf u} {D t} & = & - \nabla w + \mathbf g \\
+\nabla \cdot \mathbf u & = & 0
+\end{cases}$
+where:
+:$\mathbf u$ denotes the [[Definition:Flow Velocity|flow velocity]] [[Definition:Vector|vector]], with components in an $N$-dimensional space $u_1, u_2, \dots, u_N$
+:$\dfrac D {D t}$ denotes the [[Definition:Material Defivative|material derivative]] in time
+:$\cdot$ denotes the [[Definition:Dot Product|dot product]]
+:$\nabla$ denotes the [[Definition:Nabla Operator|nabla operator]], used to represent the specific [[Definition:Thermodynamic Work Gradient|thermodynamic work gradient]] (first equation), and the [[Definition:Flow Velocity|flow velocity]] [[Definition:Divergence|divergence]] (second equation)
+:$\mathbf u \cdot \nabla$ is the [[Definition:Convective Derivative|convective derivative]]
+:$w$ is the [[Definition:Specific Thermodynamic Work|thermodynamic work per unit mass]], the internal [[Definition:Linear Differential Equation|source term]]
+:$\mathbf g$ denotes [[Definition:Body Acceleration|body acceleration]] per unit mass acting on the continuum.
+{{stub|Absolutely no idea as to how best to present this. Leaving it as a stub for now.}}
+{{Namedfor|Leonhard Paul Euler|cat = Euler}}
+\end{theorem}<|endoftext|>
+\section{Euler-Bernoulli Beam Equation}
+Tags: Mechanics
+
+\begin{theorem}
+:$\map {\dfrac {\d^2} {\d x^2} } {E I \dfrac {\d^2 w} {\d x^2} } = q$
+{{MissingLinks|Symbols}}
+\end{theorem}
+
+\begin{proof}
+{{proof wanted}}
+{{Namedfor|Leonhard Paul Euler|name2 = Daniel Bernoulli|cat = Euler|cat2 = Bernoulli, Daniel}}
+\end{proof}<|endoftext|>
+\section{Euler Buckling Formula}
+Tags: Mechanics
+
+\begin{theorem}
+:$F = \dfrac {\pi^2 E I} {\paren {K L}^2}$
+where:
+:$F$ = maximum or critical [[Definition:Force|force]] (vertical load on column)
+:$E$ = [[Definition:Modulus of Elasticity|modulus of elasticity]]
+:$I$ = [[Definition:Area Moment of Inertia|area moment of inertia]] of the cross section of the rod
+:$L$ = unsupported length of column
+:$K$ = [[:File:ColumnEffectiveLength.png|column effective length factor]], whose value depends on the conditions of end support of the column, as follows:
+::For both ends pinned (hinged, free to rotate), $K = 1.0$
+::For both ends fixed, $K = 0.50$
+::For one end fixed and the other end pinned, $K \approx 0.699$
+::For one end fixed and the other end free to move laterally, $K = 2.0$
+:$K L$ is the effective length of the column
+\end{theorem}<|endoftext|>
+\section{Mellin Transform of Heaviside Step Function/Lemma}
+Tags: Heaviside Step Function
+
+\begin{theorem}
+Let $t \in \R$.
+Let $s \in \C$ with $\map \Re s < 0$.
+Then:
+:$\displaystyle \lim_{t \mathop \to +\infty} t^s = 0$
+\end{theorem}
+
+\begin{proof}
+Let $s = a + b i$, where $a, b \in \R$, $a<0$.
+{{begin-eqn}}
+{{eqn | l = \cmod {\lim_{t \mathop \to + \infty} t^s}
+      | r = \lim_{t \mathop \to +\infty} \cmod {t^s}
+      | c = 
+}}
+{{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{s \ln t} }
+      | c = {{Defof|Power to Complex Number}}
+}}
+{{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{\paren {a + b i} \ln t} }
+      | c = 
+}}
+{{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{a \ln t} e^{i b \ln t} }
+      | c = 
+}}
+{{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{a \ln t} } \cmod {e^{i b \ln t} }
+      | c = [[Modulus of Product]]
+}}
+{{eqn | r = \lim_{t \mathop \to +\infty} \cmod {e^{a \ln t} }
+      | c = [[Modulus of Exponential of Imaginary Number is One]]
+}}
+{{eqn | r = \lim_{t \mathop \to +\infty} \cmod {t^a}
+      | c = {{Defof|Power to Complex Number}}
+}}
+{{eqn | r = \lim_{t \mathop \to +\infty} \cmod {\dfrac 1 {t^{-a} } }
+      | c = 
+}}
+{{eqn | r = 0
+      | c = [[Sequence of Powers of Reciprocals is Null Sequence/Real Index|Sequence of Powers of Reciprocals is Null Sequence]]
+}}
+{{eqn | ll= \leadsto
+      | l = \lim_{t \mathop \to +\infty} t^s
+      | r = 0
+      | c = 
+}}
+{{end-eqn}}
+{{qed}}
+[[Category:Heaviside Step Function]]
+fftjekx69m6362u7vuu4botdlc11iua
+\end{proof}<|endoftext|>
+\section{Up-Complete Lower Bounded Join Semilattice is Complete}
+Tags: Complete Lattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Up-Complete|up-complete]] [[Definition:Bounded Below Set|lower bounded]] [[Definition:Join Semilattice|join semillattice]].
+Then $\left({S, \preceq}\right)$ is [[Definition:Complete Lattice|complete]].
+\end{theorem}
+
+\begin{proof}
+Let $X$ be a [[Definition:Subset|subset]] of $S$.
+In the case when $X = \varnothing$:
+by definition of [[Definition:Bounded Below Set|lower bounded]]:
+:$\exists L \in S: L$ is [[Definition:Lower Bound of Set|lower bound]] for $S$.
+By definition of [[Definition:Empty Set|empty set]]:
+:$L$ is [[Definition:Upper Bound of Set|upper bound]] for $X$.
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$\forall x \in S: x$ is [[Definition:Upper Bound of Set|upper bound]] for $X \implies L \preceq x$
+Thus by definition
+:$L$ is a [[Definition:Supremum of Set|supremum]] of $X$.
+Thus:
+:$X$ admint a [[Definition:Supremum of Set|supremum]].
+{{qed|lemma}}
+In the case when $X \ne \varnothing$:
+Define
+:$Y := \left\{ {\sup A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$
+where ${\it Fin}\left({X}\right)$ denotes the [[Definition:Set of Sets|set]] of all [[Definition:Finite Set|finite]] [[Definition:Subset|subsets]] of $X$.
+By [[Existence of Non-Empty Finite Suprema in Join Semilattice]]
+:all [[Definition:Supremum of Set|suprema]] in $Y$ exist,
+By definition of [[Definition:Non-Empty Set|non-empty set]]:
+:$Y$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
+We will prove that
+:$Y$ is [[Definition:Directed Subset|directed]]
+Let $x, y \in Y$.
+By definition of $Y$:
+:$\exists A \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: x = \sup A$
+and
+:$\exists B \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: y = \sup B$
+By [[Finite Union of Finite Sets is Finite]]:
+:$A \cup B$ is [[Definition:Finite Set|finite]]
+:$A \cup B \ne \varnothing$
+By [[Union is Smallest Superset]]:
+:$A \cup B \subseteq X$
+By definition of $Y$:
+:$\sup\left({A \cup B}\right) \in Y$
+By [[Set is Subset of Union]]:
+:$A \subseteq A \cup B$ and $B \subseteq A \cup B$
+Thus by [[Supremum of Subset]]:
+:$x \preceq \sup\left({A \cup B}\right)$ and $y \preceq \sup\left({A \cup B}\right)$
+Thus by definition:
+:$Y$ is [[Definition:Directed Subset|directed]].
+By definition [[Definition:Up-Complete|up-complete]]:
+:$Y$ admits a [[Definition:Supremum of Set|supremum]]
+By definition of [[Definition:Supremum of Set|supremum]]
+:$\sup Y$ is [[Definition:Upper Bound of Set|upper bound]] for $Y$
+We will prove that
+:$X \subseteq Y$
+Let $x \in X$.
+By definitions of [[Definition:Subset|subset]] and [[Definition:Singleton|singleton]]:
+:$\left\{ {x}\right\} \subseteq X$
+:$\left\{ {x}\right\}$ is [[Definition:Finite Set|finite]]
+:$\left\{ {x}\right\} \ne \varnothing$
+By definition of $Y$:
+:$\sup\left\{ {x}\right\} \in Y$
+Thus by [[Supremum of Singleton]]:
+:$x \in Y$
+By [[Upper Bound is Upper Bound for Subset]]:
+:$\sup Y$ is [[Definition:Upper Bound of Set|upper bound]] for $X$
+We will prove that
+:$\forall x \in S: x$ is [[Definition:Upper Bound of Set|upper bound]] for $X \implies \sup Y \preceq x$
+Let $x \in S$ such that
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $X$
+We will prove as sublemma that
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $Y$
+Let $y \in Y$.
+By definition of $Y$:
+:$\exists A \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: y = \sup A$
+By definition of ${\it Fin}$:
+:$A \subseteq X$
+By [[Upper Bound is Upper Bound for Subset]]
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $A$
+Thus by definition of [[Definition:Supremum of Set|supremum]]:
+:$y \preceq x$
+Thus by definition
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $Y$
+This ends the proof of sublemma.
+Thus by definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup Y \preceq x$
+This ends the proof of lemma.
+By definition
+:$\sup Y$ is a [[Definition:Supremum of Set|supremum]] of $X$
+and thus:
+:$X$ admits a [[Definition:Supremum of Set|supremum]].
+{{qed|lemma}}
+Thus result follows by [[Lattice is Complete iff it Admits All Suprema]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Gregory Series}
+Tags: Tangent Function
+
+\begin{theorem}
+For $-\dfrac \pi 4 \le \theta \le \dfrac \pi 4$:
+:$\theta = \tan \theta - \dfrac 1 3 \tan^3 \theta + \dfrac 1 5 \tan^5 \theta - \ldots$
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = 1
+      | r = \frac {\sec^2 \theta} {\sec^2 \theta}
+      | c = 
+}}
+{{eqn | l = 
+      | r = \sec^2 \theta \times \frac 1 {1 - \paren {-\tan^2 \theta} }
+      | c = 
+}}
+{{eqn | l = 
+      | r = \sec^2 \theta \times \sum_{n \mathop = 0}^\infty \paren {-\tan^2 \theta}^n
+      | c = [[Sum of Infinite Geometric Sequence]]
+}}
+{{eqn | l = 
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \sec^2 \theta \tan^{2 n} \theta
+      | c =
+}}
+{{end-eqn}}
+By the [[Nth Root Test|root test]] the [[Definition:Radius of Convergence|radius of convergence]] is $-\dfrac \pi 4 \le \theta \le \dfrac \pi 4$.
+{{begin-eqn}}
+{{eqn | l = \int \paren {1} \rd \theta
+      | r = \int \paren {\sum_{n \mathop = 0} ^ \infty \paren {-1} ^ n \sec ^ 2 \theta \tan ^ {2 n} \theta} \rd \theta
+      | c = [[Definition:Integration|Integrating]] both sides
+}}
+{{eqn | l = \int \rd \theta
+      | r = \sum_{n \mathop = 0}^\infty \paren {-1} ^ n \int  \paren { \sec ^ 2 \theta \tan ^ {2 n} \theta} \rd \theta
+      | c = 
+}}
+{{eqn | l = \theta
+      | r = \sum_{n \mathop = 0} ^ \infty \frac {\paren {-1} ^ n}{2 n + 1} \tan ^ {2 n + 1} \theta
+      | c = [[Primitive of Power of Tangent of a x by Square of Secant of a x]]
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Differential Form of Arc Length on Curved Surface}
+Tags: Differential Geometry
+
+\begin{theorem}
+The [[Definition:First Fundamental Form|first fundamental form]] for the element of [[Definition:Arc Length|arc length]] on a [[Definition:Curve|curved]] [[Definition:Surface|surface]] is given by:
+:$\d s^2 = E \rd u^2 + 2 F \d u \rd v + G \rd v^2$
+where $E$, $F$ and $G$ are the [[Definition:Coefficient|coefficients]] of the [[Definition:First Fundamental Form|first fundamental form]].
+\end{theorem}<|endoftext|>
+\section{Lattice is Complete iff it Admits All Suprema}
+Tags: Complete Lattices
+
+\begin{theorem}
+Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
+Then $\left({S, \preceq}\right)$ is a [[Definition:Complete Lattice|complete lattice]] {{iff}}
+:$\forall X \subseteq S: X$ admits a [[Definition:Supremum of Set|supremum]].
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Assume that
+$\left({S, \preceq}\right)$ is a [[Definition:Complete Lattice|complete lattice]].
+Thus by definition of [[Definition:Complete Lattice|complete lattice]]:
+:$\forall X \subseteq S: X$ admits a [[Definition:Supremum of Set|supremum]].
+{{qed|lemma}}
+=== Necessary Condition ===
+Assume that
+:$\forall X \subseteq S: X$ admits a [[Definition:Supremum of Set|supremum]].
+We will prove that
+:$\forall X \subseteq S: X$ admits both a [[Definition:Supremum of Set|supremum]] and an [[Definition:Infimum of Set|infimum]].
+Let $X \subseteq S$.
+Thus by assumption
+:$X$ admits a [[Definition:Supremum of Set|supremum]].
+Define
+:$Y := \left\{ {s \in S: s}\right.$ id [[Definition:Lower Bound of Set|lower bound]] for $\left.{X}\right\}$
+By assumption
+:$Y$ admits a [[Definition:Supremum of Set|supremum]]
+We will prove that
+:$\sup Y$ is [[Definition:Lower Bound of Set|lower bound]] for $X$
+Let $x \in X$.
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$x$ is [[Definition:Upper Bound of Set|upper bound]] for $Y$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup Y \preceq x$
+thus by definition
+:$\sup Y$ is [[Definition:Lower Bound of Set|lower bound]] for $X$
+We will prove that
+:$\forall s \in S: s$ is [[Definition:Lower Bound of Set|lower bound]] for $X \implies s \preceq \sup Y$
+Let $s \in S$ such that
+:$s$ is [[Definition:Lower Bound of Set|lower bound]] for $X$.
+By definition of $Y$:
+:$s \in Y$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\sup Y$ is [[Definition:Upper Bound of Set|upper bound]] for $Y$.
+Thus by definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$s \preceq \sup Y$
+Thus by definition
+:$X$ admits an [[Definition:Infimum of Set|infimum]]
+By definition
+:$\left({S, \preceq}\right)$ is a [[Definition:Lattice|lattice]].
+Thus the result follows by definition of [[Definition:Complete Lattice|complete lattice]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Theorema Egregium}
+Tags: Differential Geometry, Named Theorems
+
+\begin{theorem}
+The [[Definition:Gaussian Curvature|Gaussian curvature]] of a [[Definition:Surface|surface]] does not change if one bends the [[Definition:Surface|surface]] without stretching it.
+That is, [[Definition:Gaussian Curvature|Gaussian curvature]] can be determined entirely by measuring [[Definition:Angle|angles]], [[Definition:Distance (Linear Measure)|distances]] and their [[Definition:Rate of Change|rates]] on the [[Definition:Surface|surface]] itself, without further reference to the particular way in which the [[Definition:Surface|surface]] is embedded in the ambient [[Definition:Dimension (Geometry)|$3$-dimensional]] [[Definition:Euclidean Space|Euclidean space]].
+Thus the [[Definition:Gaussian Curvature|Gaussian curvature]] is an intrinsic invariant of a [[Definition:Surface|surface]].
+\end{theorem}<|endoftext|>
+\section{Gauss-Bonnet Theorem}
+Tags: Differential Geometry
+
+\begin{theorem}
+Let $M$ be a [[Definition:Compact Space|compact]] [[Definition:Dimension (Geometry)|$2$-dimensional]] [[Definition:Riemannian Manifold|Riemannian manifold]] with [[Definition:Boundary (Topology)|boundary]] $\partial M$.
+Let $\Kappa$ be the [[Definition:Gaussian Curvature|Gaussian curvature]] of $M$.
+Let $k_g$ be the [[Definition:Geodesic Curvature|geodesic curvature]] of $\partial M$.
+Then :
+:$\displaystyle \int_M \kappa \, \mathrm d A + \int_{\partial M} k_g \, \mathrm d s = 2 \pi \chi\left({M}\right)$
+where:
+:$\mathrm d A$ is the [[Definition:Area Element|element of area]] of the [[Definition:Surface|surface]]
+:$\mathrm d s$ is the [[Definition:Line Element|line element]] along $\partial M$
+:$\chi\left({M}\right)$ is the [[Definition:Euler Characteristic|Euler characteristic]] of $M$.
+\end{theorem}
+
+\begin{proof}
+{{ProofWanted}}
+{{Namedfor|Carl Friedrich Gauss|name2 = Pierre Ossian Bonnet|cat = Gauss|cat2 = Bonnet}}
+\end{proof}<|endoftext|>
+\section{Unique Factorization Theorem for Gaussian Integers}
+Tags: Algebraic Number Theory, Gaussian Primes
+
+\begin{theorem}
+For every [[Definition:Gaussian Integer|Gaussian integer]] $n$ such that $\left\vert{n}\right\vert > 1$, $n$ can be expressed as the [[Definition:Complex Multiplication|product]] of one or more [[Definition:Gaussian Prime|Gaussian primes]], uniquely up to the order in which they appear.
+\end{theorem}<|endoftext|>
+\section{Dirichlet's Principle (Harmonic Functions)}
+Tags: Potential Theory
+
+\begin{theorem}
+Let the [[Definition:Function|function]] $\map u x$ be the [[Definition:Particular Solution to Differential Equation|particular solution]] to [[Definition:Poisson's Equation|Poisson's equation]]:
+:$\Delta u + f = 0$
+on a [[Definition:Domain of Mapping|domain]] $\Omega$ of $\R^n$ with [[Definition:Boundary Condition|boundary condition]]:
+:$u = g$ on $\partial \Omega$
+Then $u$ can be obtained as the minimizer of the [[Definition:Dirichlet's Energy|Dirichlet's energy]]:
+:$\displaystyle E \sqbrk {\map v x} = \int_\Omega \paren {\frac 1 2 \cmod {\nabla v}^2 - v f} \rd x$
+amongst all [[Definition:Differentiable Function|twice differentiable functions]] $v$ such that $v = g$ on $\partial \Omega$ .
+This result holds provided that there exists at least one [[Definition:Function|function]] which makes the [[Definition:Dirichlet Integral|Dirichlet integral]] finite.
+\end{theorem}<|endoftext|>
+\section{Galois Connection is Expressed by Minimum}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be [[Definition:Ordered Set|ordered sets]].
+Let $g: S \to T$, $d: T \to S$ be [[Definition:Mapping|mappings]].
+Then $\tuple {g, d}$ is a [[Definition:Galois Connection|Galois connection]] {{iff}}:
+:$g$ is an [[Definition:Increasing Mapping|increasing mapping]] and
+::$\forall t \in T: \map d t = \min \set {g^{-1} \sqbrk {t^\succsim} }$
+where
+:$\min$ denotes the [[Definition:Smallest Element of Subset|minimum]]
+:$g^{-1} \sqbrk {t^\succsim}$ denotes the [[Definition:Image of Subset under Mapping|image]] of $t^\succsim$ under relation $g^{-1}$
+:$t^\succsim$ denotes the [[Definition:Upper Closure/Element|upper closure]] of $t$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\left({g, d}\right)$ be a [[Definition:Galois Connection|Galois connection]].
+Thus by definition of [[Definition:Galois Connection|Galois connection]]:
+:$g$ is an [[Definition:Increasing Mapping|increasing mapping]]
+Let $t \in T$.
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$d\left({t}\right) \preceq d\left({t}\right)$
+By definition of [[Definition:Galois Connection|Galois connection]]:
+:$t \precsim g\left({d\left({t}\right)}\right)$
+By definition of [[Definition:Upper Closure/Element|upper closure]]:
+:$g\left({d\left({t}\right)}\right) \in t^\succsim$
+By definition of [[Definition:Image of Subset under Relation|image of set under relation]]:
+:$d\left({t}\right) \in g^{-1}\left[{t^\succsim}\right]$
+By definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$\forall s \in S: s$ is [[Definition:Lower Bound of Set|lower bound]] for $g^{-1}\left[{t^\succsim}\right] \implies s \preceq d\left({t}\right)$
+We will prove that
+:$d\left({t}\right)$ is [[Definition:Lower Bound of Set|lower bound]] for $g^{-1}\left[{t^\succsim}\right]$
+Let $s \in g^{-1}\left[{t^\succsim}\right]$.
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$g\left({s}\right) \in t^\succsim$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$t \precsim g\left({s}\right)$
+Thus by definition of [[Definition:Galois Connection|Galois connection]]:
+:$d\left({t}\right) \preceq s$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$g^{-1}\left[{t^\succsim}\right]$ admits an [[Definition:Infimum of Set|infimum]]
+and
+:$\inf\left({g^{-1}\left[{t^\succsim}\right]}\right) = d\left({t}\right)$
+Thus:
+:$d\left({t}\right) = \min\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $g: S \to T$ be an [[Definition:Increasing Mapping|increasing mapping]].
+Let $\forall t \in T: d\left({t}\right) = \min\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+Thus:
+:$g$ is [[Definition:Increasing Mapping|increasing mapping]].
+We will prove that
+:$d$ is [[Definition:Increasing Mapping|increasing mapping]].
+Let $x, y \in T$ such that
+:$x \precsim y$
+By [[Upper Closure is Decreasing]]:
+:$y^\succsim \subseteq x^\succsim$
+By [[Image of Subset under Relation is Subset of Image/Corollary 3]]:
+:$g^{-1}\left[{y^\succsim}\right] \subseteq g^{-1}\left[{x^\succsim}\right]$
+By assumption
+:$d\left({x}\right) = \min\left({g^{-1}\left[{x^\succsim}\right]}\right) = \inf\left({g^{-1}\left[{x^\succsim}\right]}\right)$
+and
+:$d\left({y}\right) = \min\left({g^{-1}\left[{y^\succsim}\right]}\right) = \inf\left({g^{-1}\left[{y^\succsim}\right]}\right)$
+By [[Infimum of Subset]]:
+:$d\left({x}\right) \preceq d\left({y}\right)$
+Thus by definition:
+:$d$ is an [[Definition:Increasing Mapping|increasing mapping]].
+We will prove that
+:$\forall s \in S, t \in T:  t \precsim g\left({s}\right) \iff d\left({t}\right) \preceq s$
+Let $s \in S, t \in T$.
+First implication:
+Let $t \precsim g\left({s}\right)$.
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$g\left({s}\right) \in t^\succsim$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$s \in g^{-1}\left[{t^\succsim}\right]$
+By assumption
+:$d\left({t}\right) = \min\left({g^{-1}\left[{t^\succsim}\right]}\right) = \inf\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+By definition of [[Definition:Infimum of Set|infimum]]:
+:$d\left({t}\right)$ is [[Definition:Lower Bound of Set|lower bound]] for $g^{-1}\left[{t^\succsim}\right]$
+Thus by definition of [[Definition:Lower Bound of Set|lower bound]]:
+:$d\left({t}\right) \preceq s$
+Second implication:
+Let $d\left({t}\right) \preceq s$
+By assumption
+:$d\left({t}\right) = \min\left({g^{-1}\left[{t^\succsim}\right]}\right)$
+By [[Definition:Smallest/Ordered Set/Subset|smallest element of set]]:
+:$d\left({t}\right) \in g^{-1}\left[{t^\succsim}\right]$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$g\left({d\left({t}\right)}\right) \in t^\succsim$
+By definition of [[Definition:Upper Closure/Element|upper closure of element]]:
+:$t \precsim g\left({d\left({t}\right)}\right)$
+By definition of [[Definition:Increasing Mapping|increasing mapping]]:
+:$g\left({d\left({t}\right)}\right) \precsim g\left({s}\right)$
+Thus by definition of [[Definition:Transitivity|transitivity]]:
+:$t \precsim g\left({s}\right)$
+Thus by definition:
+:$\left({g, d}\right)$ is [[Definition:Galois Connection|Galois connection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Form for Pentagonal Numbers}
+Tags: Pentagonal Numbers, Closed Forms
+
+\begin{theorem}
+The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Pentagonal Number|pentagonal number]] is:
+:$P_n = \dfrac {n \paren {3 n - 1} } 2$
+\end{theorem}
+
+\begin{proof}
+[[Definition:Pentagonal Number|Pentagonal numbers]] are [[Definition:Polygonal Number|$k$-gonal numbers]] where $k = 5$.
+From [[Closed Form for Polygonal Numbers]] we have that:
+:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = P_n
+      | r = \frac n 2 \paren {\paren {5 - 2} n - 5 + 4}
+      | c = [[Closed Form for Polygonal Numbers]]
+}}
+{{eqn | r = \dfrac {n \paren {3 n - 1} } 2
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Closed Form for Hexagonal Numbers}
+Tags: Hexagonal Numbers, Closed Forms
+
+\begin{theorem}
+The [[Definition:Closed-Form Expression|closed-form expression]] for the $n$th [[Definition:Hexagonal Number|hexagonal number]] is:
+:$H_n = n \paren {2 n - 1}$
+\end{theorem}
+
+\begin{proof}
+[[Definition:Hexagonal Number|Hexagonal numbers]] are [[Definition:Polygonal Number|$k$-gonal numbers]] where $k = 6$.
+From [[Closed Form for Polygonal Numbers]] we have that:
+:$\map P {k, n} = \dfrac n 2 \paren {\paren {k - 2} n - k + 4}$
+Hence:
+{{begin-eqn}}
+{{eqn | l = H_n
+      | r = \frac n 2 \paren {\paren {6 - 2} n - 6 + 4}
+      | c = [[Closed Form for Polygonal Numbers]]
+}}
+{{eqn | r = n \paren {2 n - 1}
+      | c = 
+}}
+{{end-eqn}}
+Hence the result.
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Sum of Even Integers is Even}
+Tags: Even Integers, Euclidean Number Theory, Sum of Even Integers is Even
+
+\begin{theorem}
+The [[Definition:Integer Addition|sum]] of any [[Definition:Finite Set|finite number]] of [[Definition:Even Integer|even integers]] is itself [[Definition:Even Integer|even]].
+\end{theorem}
+
+\begin{proof}
+Let $S = \set {r_1, r_2, \ldots, r_n}$ be a [[Definition:Set|set]] of $n$ [[Definition:Even Integer|even numbers]].
+By definition of [[Definition:Even Integer|even number]], this can be expressed as:
+:$S = \set {2 s_1, 2 s_2, \ldots, 2 s_n}$
+where:
+:$\forall k \in \closedint 1 n: r_k = 2 s_k$
+Then:
+{{begin-eqn}}
+{{eqn | l = \sum_{k \mathop = 1}^n r_k
+      | r = \sum_{k \mathop = 1}^n 2 s_k
+      | c = 
+}}
+{{eqn | r = 2 \sum_{k \mathop = 1}^n s_k
+      | c = 
+}}
+{{end-eqn}}
+Thus, by definition, $\displaystyle \sum_{k \mathop = 1}^n r_k$ is [[Definition:Even Integer|even]].
+{{qed}}
+\end{proof}
+
+\begin{proof}
+Proof by [[Principle of Mathematical Induction|induction]]:
+For all $n \in \N$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
+:The sum of $n$ [[Definition:Even Integer|even integers]] is an [[Definition:Even Integer|even integer]].
+$P(1)$ is trivially true, as this just says:
+:The sum of $1$ [[Definition:Even Integer|even integers]] is an [[Definition:Even Integer|even integer]].
+The sum of $0$ [[Definition:Even Integer|even integers]] is understood, from the definition of a [[Definition:Vacuous Summation|vacuous summation]], to be $0$, which is [[Definition:Even Integer|even]].
+So $P(0)$ is also true.
+=== Basis for the Induction ===
+$P(2)$ is the case:
+:The [[Definition:Integer Addition|sum]] of any two [[Definition:Even Integer|even integers]] is itself [[Definition:Even Integer|even]].
+Consider two [[Definition:Even Integer|even integers]] $x$ and $y$.
+Since they are [[Definition:Even Integer|even]], they can be written as $x = 2a$ and $y = 2b$ respectively for [[Definition:Integer|integers]] $a$ and $b$.
+Therefore, the sum is:
+:$x + y = 2 a + 2 b = 2 \left({a + b}\right)$
+Thus $x + y$ has $2$ as a [[Definition:Divisor of Integer|divisor]] and is [[Definition:Even Integer|even]] by definition.
+This is our [[Principle of Mathematical Induction#Basis for the Induction|basis for the induction]].
+{{qed|lemma}}
+=== Induction Hypothesis ===
+Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.
+So this is our [[Principle of Mathematical Induction#Induction Hypothesis|induction hypothesis]]:
+:The [[Definition:Integer Addition|sum]] of any $k$ [[Definition:Even Integer|even integers]] is itself even.
+Then we need to show:
+:The [[Definition:Integer Addition|sum]] of any $k+1$ [[Definition:Even Integer|even integers]] is itself even.
+=== Induction Step ===
+This is our [[Principle of Mathematical Induction#Induction Step|induction step]]:
+Consider the [[Definition:Integer Addition|sum]] of any $k + 1$ [[Definition:Even Integer|even integers]].
+This is the [[Definition:Integer Addition|sum]] of:
+: $k$ [[Definition:Even Integer|even integers]] (which is even by the [[Sum of Even Integers is Even#Induction Hypothesis|induction hypothesis]])
+and:
+: another [[Definition:Even Integer|even integer]].
+That is, it is the [[Definition:Integer Addition|sum]] of two [[Definition:Even Integer|even integers]].
+By the [[Sum of Even Integers is Even#Basis for the Induction|basis for the induction]], this is also [[Definition:Even Integer|even]].
+So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the [[Principle of Mathematical Induction]].
+That is:
+:The [[Definition:Integer Addition|sum]] of any [[Definition:Finite Set|finite number]] of [[Definition:Even Integer|even integers]] is itself [[Definition:Even Integer|even]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Galois Connection is Expressed by Maximum}
+Tags: Galois Connections
+
+\begin{theorem}
+Let $\struct {S, \preceq}$, $\struct {T, \precsim}$ be [[Definition:Ordered Set|ordered sets]].
+Let $g: S \to T$, $d: T \to S$ be [[Definition:Mapping|mappings]].
+Then $\tuple {g, d}$ is a [[Definition:Galois Connection|Galois connection]] {{iff}}
+:$d$ is an [[Definition:Increasing Mapping|increasing mapping]] and:
+::$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
+where:
+:$\max$ denotes the [[Definition:Greatest Element of Subset|maximum]]
+:$d^{-1} \sqbrk {s^\preceq}$ denotes the [[Definition:Image of Subset under Mapping|image]] of $s^\preceq$ under relation $d^{-1}$
+:$s^\preceq$ denotes the [[Definition:Lower Closure/Element|lower closure]] of $t$
+\end{theorem}
+
+\begin{proof}
+=== Sufficient Condition ===
+Let $\tuple {g, d}$ be a [[Definition:Galois Connection|Galois connection]].
+Thus by definition of [[Definition:Galois Connection|Galois connection]]:
+:$d$ is an [[Definition:Increasing Mapping|increasing mapping]].
+Let $s \in S$.
+By definition of [[Definition:Reflexivity|reflexivity]]:
+:$\map g s \preceq \map g s$
+By definition of [[Definition:Galois Connection|Galois connection]]:
+:$\map d {\map g s} \preceq s$
+By definition of [[Definition:Lower Closure/Element|lower closure]]:
+:$\map d {\map g s} \in s^\preceq$
+By definition of [[Definition:Image of Subset under Relation|image of set under relation]]:
+:$\map g s \in d^{-1} \sqbrk {s^\preceq}$
+By definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$\forall t \in T: t$ is an [[Definition:Upper Bound of Set|upper bound]] for $d^{-1} \sqbrk {s^\preceq} \implies \map g s \precsim t$
+We will prove that:
+:$\map g s$ is an [[Definition:Upper Bound of Set|upper bound]] for $d^{-1} \sqbrk {s^\preceq}$
+Let $t \in d^{-1} \sqbrk {s^\preceq}$.
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$\map d t \in s^\preceq$
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$\map d t \preceq s$
+Thus by definition of [[Definition:Galois Connection|Galois connection]]:
+:$t \precsim \map g s$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$d^{-1} \sqbrk {s^\preceq}$ admits a [[Definition:Supremum of Set|supremum]]
+and
+:$\map \sup {d^{-1} \sqbrk {s^\preceq} } = \map g s$
+Thus:
+:$\map g s = \map \min {d^{-1} \sqbrk {s^\preceq} }$
+{{qed|lemma}}
+=== Necessary Condition ===
+Let $d: T \to S$ be an [[Definition:Increasing Mapping|increasing mapping]].
+Let:
+:$\forall s \in S: \map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
+We will prove that:
+:$g$ is an [[Definition:Increasing Mapping|increasing mapping]].
+Let $x, y \in S$ such that:
+:$x \preceq y$
+By [[Lower Closure is Increasing]]:
+:$x^\preceq \subseteq y^\preceq$
+By [[Image of Subset under Relation is Subset of Image/Corollary 3]]:
+:$d^{-1} \sqbrk {x^\preceq} \subseteq d^{-1} \sqbrk {y^\preceq}$
+By assumption:
+:$\map g s = \map \max {d^{-1} \sqbrk {x^\preceq} } = \map \sup {d^{-1} \sqbrk {x^\preceq} }$
+and:
+:$\map g y = \map \max {d^{-1} \sqbrk {y^\preceq} } = \map \sup {d^{-1} \sqbrk {y^\preceq} }$
+By [[Supremum of Subset]]:
+:$\map g s \preceq \map g y$
+Thus by definition:
+:$g$ is an [[Definition:Increasing Mapping|increasing mapping]].
+Thus:
+:$d$ is an [[Definition:Increasing Mapping|increasing mapping]].
+We will prove that:
+:$\forall s \in S, t \in T:  t \precsim \map g s \iff \map d t \preceq s$
+Let $s \in S, t \in T$.
+Second implication:
+Let $\map d t \preceq s$.
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$\map d t \in s^\preceq$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$t \in d^{-1} \sqbrk {s^\preceq}$
+By assumption:
+:$\map g s = \map \max {d^{-1} \sqbrk {s^\preceq} } = \map \sup {d^{-1} \sqbrk {s^\preceq} }$
+By definition of [[Definition:Supremum of Set|supremum]]:
+:$\map g s$ is an [[Definition:Upper Bound of Set|upper bound]] for $d^{-1} \sqbrk {s^\preceq}$
+Thus by definition of [[Definition:Upper Bound of Set|upper bound]]:
+:$t \precsim \map g s$
+First implication:
+Let $t \precsim \map g s$
+By assumption:
+:$\map g s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
+By [[Definition:Greatest Element of Subset|greatest element of set]]:
+:$\map g s \in d^{-1} \sqbrk {s^\preceq}$
+By definition of [[Definition:Image of Subset under Relation|image of set]]:
+:$\map d {\map g s} \in s^\preceq$
+By definition of [[Definition:Lower Closure/Element|lower closure of element]]:
+:$\map d {\map g s} \preceq s$
+By definition of [[Definition:Increasing Mapping|increasing mapping]]:
+:$\map d t \preceq \map d {\map g s}$
+Thus by definition of [[Definition:Transitivity|transitivity]]:
+:$\map d t \preceq s$
+Thus by definition:
+:$\tuple {g, d}$ is a [[Definition:Galois Connection|Galois connection]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Compass and Straightedge Construction for Regular Heptagon does not exist}
+Tags: Regular Polygons
+
+\begin{theorem}
+There exists no [[Definition:Compass and Straightedge Construction|compass and straightedge construction]] for the [[Definition:Regular Heptagon|regular heptagon]].
+\end{theorem}
+
+\begin{proof}
+By definition, the [[Definition:Regular Heptagon|regular heptagon]] has $7$ sides.
+$7$ is a [[Definition:Prime Number|prime number]] which is not a [[Definition:Fermat Prime|fermat prime]].
+The result follows [[Construction of Regular Prime p-Gon Exists iff p is Fermat Prime|Construction of Regular Prime $p$-Gon Exists iff $p$ is Fermat Prime]].
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Integer to Rational Power is Irrational iff not Integer or Reciprocal}
+Tags: Number Theory, Irrational Numbers
+
+\begin{theorem}
+Let $m$ be a [[Definition:Rational Number|rational number]].
+Let $n$ be a [[Definition:Positive Integer|positive integer]].
+Then $n^m$ is an [[Definition:Irrational Number|irrational number]] {{iff}} $n^{\size m}$ is not an [[Definition:Integer|integer]].
+\end{theorem}
+
+\begin{proof}
+=== Necessary Condition ===
+Let $n^{\size m} \notin \Z$.
+We have that:
+:$n^m = n^{u/v}$ for some $u \in\Z$ and $v \in \Z_{\ne 0}$
+Then it follows from the definition of a [[Definition:Rational Power|rational power]] and the [[Existence of Positive Root of Positive Real Number|existence]] of a [[Definition:Real Number|real]] [[Definition:Root (Analysis)|$v$th root of $n^u$]] that:
+:$n^m = n^{u/v} = \paren {n^u}^{1/v} \in \R$
+{{AimForCont}} $n^m \in \Q$.
+We have that:
+:$\size m = 0 \implies n^{\size m} \in \Z$
+It follows that:
+:$\size m > 0$
+Then:
+:$\size m > 0$
+:$m \in \Q$
+:$n^{\size m} > 0$
+:$n^m \in \Q$
+imply that
+:$n^{\size m} = n^{p/q} = \dfrac r s$
+for some $\tuple {p, q, r, s} \in \Z_{>0}^4$ where $r$ and $s$ have no [[Definition:Common Divisor of Integers|common]] [[Definition:Prime Factor|prime factors]].
+Raising both sides of $n^{p/q} = \dfrac r s$ to the power of $q$ yields:
+:$n^p = \dfrac {r^q} {s^q}$
+We have that:
+:$\dfrac {r^q} {s^q} = n^p \in \Z$
+Hence $r^q$ is [[Definition:Divisor of Integer|divisible]] by $s^q$.
+By the [[Fundamental Theorem of Arithmetic]], $r^q$ and $s^q$ have the same [[Definition:Prime Factor|prime factors]] as $r$ and $s$ respectively.
+But we have that:
+:$s^q \divides r^q$
+and:
+:$s \ne 1$
+imply that $r$ and $s$ have a [[Definition:Common Divisor of Integers|common]] [[Definition:Prime Factor|prime factor]].
+But this would contradict the fact that $r$ and $s$ have no [[Definition:Common Divisor of Integers|common]] [[Definition:Prime Factor|prime factors]].
+Therefore $s = 1$.
+Then:
+:$n^{\size m} = \dfrac r s$
+and:
+:$s = 1$
+imply that:
+:$r = n^{\size m} \notin \Z$
+From this [[Definition:Contradiction|contradiction]] it follows that:
+:$n^m \notin \Q$
+and since $n^m \in \R$:
+:$n^m \in \R \setminus \Q$
+That is, $n^m$ is an [[Definition:Irrational Number|irrational number]].
+{{qed|lemma}}
+=== Sufficient Condition ===
+Now let $n^m$ be an [[Definition:Irrational Number|irrational number]].
+That is:
+:$n^m \in \R \setminus \Q$
+Then because $n^{-m}$ is the [[Definition:Reciprocal|reciprocal]] of $n^m$:
+:$n^{-m} \in \R \setminus \Q$
+So:
+:$n^{\size m} \in \R \setminus \Q$
+and so since $\Z \subseteq \Q$:
+:$n^{\size m} \notin \Z$
+{{qed}}
+[[Category:Number Theory]]
+[[Category:Irrational Numbers]]
+na3dwtd25a4dmxxztwcd54gbp631ts3
+\end{proof}<|endoftext|>
+\section{Tangent to Archimedean Spiral at Point}
+Tags: Archimedean Spirals
+
+\begin{theorem}
+{{stub}}
+:[[File:ArchimedeanSpiralTangent.png|500px]]
+\end{theorem}<|endoftext|>
+\section{Area Enclosed by First Turn of Archimedean Spiral}
+Tags: Archimedean Spirals
+
+\begin{theorem}
+Let $S$ be the [[Definition:Archimedean Spiral|Archimedean spiral]] defined by the equation:
+:$r = a \theta$
+The [[Definition:Area|area]] $\mathcal A$ enclosed by the first turn of $S$ and the [[Definition:Polar Axis|polar axis]] is given by:
+:$\mathcal A = \dfrac {4 \pi^3 a^2} 3$
+:[[File:ArchimedeanSpiralArea.png|500px]]
+\end{theorem}
+
+\begin{proof}
+{{begin-eqn}}
+{{eqn | l = \mathcal A
+      | r = \int_0^{2 \pi} \frac {\left({a \theta}\right)^2} 2 \, \mathrm d \theta
+      | c = [[Area between Radii and Curve in Polar Coordinates]]
+}}
+{{eqn | r = \left[{\frac {a^2 \theta^3} 6}\right]_0^{2 \pi}
+      | c = [[Primitive of Power]]
+}}
+{{eqn | r = \frac {\left({2 \pi}\right)^3 a^2 \theta^3} 6
+      | c = 
+}}
+{{eqn | r = \frac {4 \pi^3 a^2} 3
+      | c = after simplification
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Trisecting the Angle/Archimedean Spiral}
+Tags: Trisecting the Angle
+
+\begin{theorem}
+Let $\alpha$ be an [[Definition:Angle|angle]] which is to be [[Definition:Trisection|trisected]].
+This can be achieved by means of an [[Definition:Archimedean Spiral|Archimedean spiral]].
+\end{theorem}
+
+\begin{proof}
+Let the equation of the [[Definition:Archimedean Spiral|Archimedean spiral]] be $r = a \theta$.
+Then:
+{{begin-eqn}}
+{{eqn | l = \angle DOB
+      | r = \frac {OD} a
+}}
+{{eqn | r = \frac {OA/3} a
+}}
+{{eqn | r = \frac 1 3 \angle AOB
+}}
+{{end-eqn}}
+{{qed}}
+\end{proof}<|endoftext|>
+\section{Squaring the Circle by Archimedean Spiral}
+Tags: Squaring the Circle, Archimedean Spirals
+
+\begin{theorem}
+Let $C$ be a [[Definition:Circle|circle]].
+It is possible to construct a [[Definition:Square (Geometry)|square]] of the same [[Definition:Area|area]] as $C$ using an [[Definition:Archimedean Spiral|Archimedean spiral]].
+\end{theorem}
\ No newline at end of file