The problem has two parts:
1. Solve the initial value problem: $$ y''+y=\sum_{j=0}^\infty \delta_{2j\pi}(t) $$ with the initial conditions: $y(0)=y'(0)=0$
2.Show that if $2n\pi<t<2(n+1)\pi$ for some integer n, then $y(t)=(n+1)sin(t)$.
I only managed to partially solve the first part. I got: $$\mathcal{L} \{y(t)\}=\frac{e^{2\pi s}}{(e^{2\pi s}-1)(s^{2}+1)}$$
I could really use some help finding the inverse Laplace transform of this and solving the second part. Thank you in advance!
We have $$\mathcal{L}\left\{y(t)\right\}=\cfrac{e^{2\pi s}}{e^{2\pi s}-1} \cfrac{1}{s^2+1} = \mathcal{L}\{x(t) \}\mathcal{L}\{\sin t\}.$$ We need to find $x(t)$, for that purpose let's observe the following: $$\cfrac{e^{2\pi s}}{e^{2\pi s}-1}=\cfrac{1}{1-e^{-2\pi s}}=\sum_{k\geq 0}e^{-2\pi ks}=\mathcal{L}\left\{ \sum_{k \geq 0} \delta_{-2k\pi}(t)\right\}.$$ We assume, of course, that $s$ is such that $e^{-2\pi s} < 1$. Note that this was what you had from the beginning. Now, with the convolution property, we have that $$y(t) = \left( \sum_{k \geq 0} \delta_{-2k\pi}(t) \right) \ast \sin t = \sum_{k \geq 0}\left( \delta_{-2k\pi}(t) \ast \sin t \right)=\sum_{k \geq 0}\sin(t-2k\pi).$$