Solving differential equation with the Dirac Delta Function I've got a differential equation to solve with the Dirac Delta Function and I'm not really sure how to handle it. I have been instructed to use the Laplace Transform as my method of solution. 
Here is the equation:
$y''+8y'+41y=δ(t-\pi)+δ(t-3\pi)$, $y(0)=1, y'(0)=0$
I have no idea where to begin here. I took the Laplace transform but at this point I'm unsure exactly how to decompose the function after I solved for $y$. 
Thanks for any help.
 A: Taking the Laplace transform is the good way!
HINTS:
$$\mathcal{L}\left(\delta(t - \pi), s\right) = e^{-\pi s}$$
$$\mathcal{L}\left(\delta(t - 3\pi), s\right) = e^{-3\pi s}$$
Can you go on? The LT of the other terms are straightforward!
A: Your homogeneous solution is $$e^{-4t}(A\cos5t+B\sin 5t).$$ To get the deltas on the right you need kinks in the solution at the given places. You get a kink at position $t=c$ as the product $y(t-c)u(t-c)$ where $y$ solves the homogeneous equation with $y(0)=0$, $y'(0)=1$ and $u$ is the unit ramp/Heaviside function. This is the case here with $$y_c(t)=\frac15e^{-4t}\sin(5(t-c))u(t-c).$$ Then
\begin{alignat}2
&&e^{4t}y_c&=\frac15\sin(5(t-c))u(t-c)\\
e^{4t}(y_c'+4y_c)&=&(e^{4t}y_c)'&=\cos(5(t-c))\,u(t-c)\\
e^{4t}(y_c''+8y_c'+16y_c)&=&(e^{4t}y_c)'' &= -5\sin(5(t-c))u(t-c)+δ(t-c)\\
\end{alignat}
$$\implies
y_c''+8y_c'+41y_c=δ(t-c)
$$
Thus a particular solution for the given right side is $$y_p(t)=\frac15\sin(5(t-\pi))u(t-\pi)+\frac15\sin(5(t-3\pi))u(t-3\pi).$$
