TITLE: Initial value problem
QUESTION [7 upvotes]: Solve the following initial value problem:
$$\frac{d^2y}{dt^2}+2\frac{dy}{dt}+5y=0; y(0)=0 \text{ and } y'(0)=2 $$
I started off with the characteristic equation which is:
$$ r^2+2r+5=0 $$
Using the quadratic equation, I found the roots to be:
$$-1+2i \text{ and } -1-2i $$
So my general equation is:
$$ y(t)=e^{-t} [c_1 \cos(2t) +c_2 \sin(2t)] $$
Using the information provided, I found $c_1=0$ and $c_2=1$, so substituting I obtain:
$$y(t)=e^{-t} \sin(2t)$$
If anyone can confirm whether I did this right would help a lot!

REPLY [3 votes]: \begin{align}
y'(t) &= e^{-t}(-\sin(2t) + 2 \cos(2t)) \\
y''(t) 
&= e^{-t}(\sin(2t) - 2 \cos(2t) -2 \cos(2t) - 4 \sin(2t)) \\
&= e^{-t}(-3 \sin(2t) - 4 \cos(2t))
\end{align}
and then
$$
y''+2 y' + 5 y =
e^{-t}((-3 + 2 + 1) \sin(2t) + (-4 + 4) \cos(2t)) = 0
$$
And $y(0) = 0$ and $y'(0) = 2$. Looks fine.