Laplace transform of a differential equation?

Question

Find the unique solution of $y''+ y = f$, $y(0) = y'(0) = 0$ with the $2\pi$ periodic function given by $f(t)=2\pi \sin(t)$.

I am having trouble setting up and starting the the question. I would be really grateful for any help or hints.

Answer 1

Hints;

Take the Laplace transform on both sides. These ones may trouble you so here they are;

$$1) \mathscr{L}_t [y''(t)](s)=s^2\mathscr{L}_t [y(t)](s)-sy(0)-y'(0)$$

$$2) \mathscr{L}_t [2\pi\sin t](s)=2\pi\mathscr{L}_t [\sin t](s)=2\pi\times\frac{1}{s^2+1}$$

$$3) \mathscr{L}_s^{-1} \left [\frac{2\pi}{(s^2+1)^2}\right ](t)=\pi\sin t-\pi t\cos t$$

$$4) \mathscr{L}_s^{-1} \left [\frac{s}{s^2+1}\right ](t)=\cos t$$