Solving a differential equation using the laplace transform involving convolution The problem is the following

The thing that puzzles me here is the integral on the right hand side, so:
How to take the laplace transform on the right hand side?
Any help to get me going would be greatly appreciated.
 A: The convolution theorem is
\begin{align}
F(s)*G(s) = \int_{0}^{t} g(t-\tau) f(\tau) d\tau
\end{align}
such that 
\begin{align}
L\{ f*g\} = L\{ \int_{0}^{t} g(t-\tau) f(\tau) d\tau \} = F(s)G(s).
\end{align}
The differential equation then becomes
\begin{align}
L\{ y^{''} + 3 y^{'} + 2y \} = L\{ H(t-\pi) \sin(2t) \} + \delta(s) f(s) - \delta(s-2) f(s).
\end{align}
Completing the transform the equation becomes
\begin{align}
(s+1)(s+2) y(s) - (s+1) y(0) - y^{'}(0) = \frac{2 e^{-\pi s}}{s^{2} + 4} + \delta(s) f(s) - \delta(s-2) f(s).
\end{align}
Since $y^{'}(0) = 0$ and $y(0) = 1$, and $f(t+2) = f(t)$ then
\begin{align}
y(s) &= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} + \delta(s) f(s) - \delta(s-2) f(s) \right] \\
&= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} + f(0) - f(2) \right] \\
&= \frac{1}{s+2} + \frac{1}{(s+1)(s+2)} \left[ \frac{2 e^{-\pi s}}{s^{2} + 4} \right] \\
\end{align}
Inversion leads to
\begin{align}
y(t) &= e^{-2t} + L^{-1}\{ \frac{1}{(s+1)(s+2)} \cdot \frac{2 e^{-\pi s}}{s^{2} + 4} \} \\
&= e^{-2t} + \frac{1}{20} H(t-\pi) \left[ 8 e^{\pi - t} - 5 e^{2(\pi -t)} - \sin(2t) - 3 \cos(2t) \right]. 
\end{align}
note: the convolution theorem is in most standard tables such as this one
http://tutorial.math.lamar.edu/pdf/Laplace_Table.pdf
A: The integral on the right-hand side can be rewritten as
$$\int_{0}^t(\delta(t-\tau)-\delta(t-2-\tau))f(\tau)d\tau=f(t)-f(t-2)=f(t),\quad 0\le t<2\quad\textrm{(and zero otherwise)}$$
So its Laplace transform is simply
$$\int_0^2f(t)e^{-st}dt$$
Also note that $h(t-\pi)\sin(2t)=h(t-\pi)\sin(2(t-\pi))$, so you can simply use the transform of $h(t)\sin(2t)$ and use the shifting property, i.e. multiply by $e^{-s\pi}$.
