Use the Laplace transform to solve the IVP for $y(t)$: \begin{align*} y''(t) + \omega^2y(t) = f(t) \qquad y(0) = y'(0) = 0 \end{align*} where $f$ is a given function on $[0,\infty)$.
My attempt to a solution:
Taking the Laplace Transform on both sides, we get: \begin{align*} \mathcal{L}\{y''(t) + \omega^2y(t)\}&=\mathcal{L}\{f(t)\}\\ \mathcal{L}\{y''(t)\}(s)+\omega^2\mathcal{L}\{y(t)\}(s)&=\mathcal{L}\{f(t)\}\\ s^2\mathcal{L}\{y(t)\}(s)-{sy(0)}-{y'(0)}+\omega^2\mathcal{L}\{y(t)\}(s)&=\mathcal{L}\{f(t)\} \tag{applying ICs}\\ (s^2+\omega^2)\mathcal{L}\{y(t)\}(s)&=\mathcal{L}\{f(t)\} \end{align*}
But I'm not sure how to proceed from here. Is it a matter of simply just moving $(s^2+\omega^2)$ to the RHS and that's it, or do I have to evaluate the RHS first?
Any tips would be appreciated!
If $f$ is not given you can only say that $y$ is the inverse Laplace transorm of $\frac {\mathcal L(f)(s)} {s^{2}+\omega^{2}}$.