TITLE: Dif. Eq. General solution is sum of homogeneous and particular solutions
QUESTION [10 upvotes]: Why is the general solution to ordinary first order differential equations a sum of homogeneous(Setting the inhomogeneous term to 0) and particular(satisfies the differential equation but not necessarily the initial conditions) solutions?
$x=x_{h}+x_{p}$
I have tried finding a proof on the Internet but without a result.
Can someone give me a not too complicated mathematical proof(If possible) to why this relation is valid?
Thank you!

REPLY [12 votes]: Let us denote the linear differential operator by $D$. We write its corresponding differential equation as $D(x)=b$, with full solution set $\Omega$.
Let $p$ be any particular solution to the equation $D(x)=b$, such that $D(p)=b$.
Now consider the set $ S=\{p+h|h    \text{ satisfies the homogeneous equation}\}$. We will show that this is equal to the solution set. We will show set inclusion both ways.
First we will show that if a general solution exists, it is in this set.
Assume that $s\in \Omega$, the full solution set to the differential equation, then consider:
$D(s-p)=D(s)-D(p)=b-b=0$.
This means that $s-p$ is a homogeneous solution, therefore we know $s=p+(s-p)\in S$. Hence $ \Omega \subseteq S$
Finally, by linearity of differentiation we have for any $p+h \in S$ :
$D(p+h)=D(p)+D(h)=b+0=b$.
So $S\subseteq \Omega$.
Thus by mutual inclusion $S=\Omega$ $\square$.