Prove that a function is solution of IVP Let $f\in \mathcal{C}^1$ and let $F$ defined by
$$F(x)=\dfrac{1}{a}\int_0^x f(t)\sin(a(x-t))dt$$
where $a\in \mathbb{R}\setminus\{0\}$. Prove that $F$ is solution of the IVP
$$y''+a^2 y'=f(x)$$
$$y(0)=y'(0)=0$$

I've used the fundamental Calculus theorem, but i can't get the request 'cause
$F'(x)=\dfrac{1}{a}f(x)\sin(a(x-x))(x')-\dfrac{1}{a}f(0)\sin(a(x-0))(0')=0$ ??
Anyone can help me? I've tried with integration by parts but i can't get the result too

Hint: $$F'(x) = \frac{1}{a} \int_0^x a f(t) \cos (a (x-t)) \, dt$$
$$F''(x) = \frac{1}{a} \int_0^x -a^2 f(t) \sin (a (x-t)) \, dt+a f(x) $$
