I wanted to solve $y''+9y=sin(3x)$, where $y(0)=0$ and $y'(0)=0$ by doing a Laplace Transform.
Taking $L[y''+9y]=L[sin(3x)]$ gives $s^{2}F(s)-sy(0)-y'(0)+9F(s)=\frac{3}{s^2+9}$
The initial values are both zero, and $F(s)$ can be factored out of the left-hand side, leaving :
$F(s)=\frac{3}{(s^2+9)^2}$ and I couldn't figure out how to inverse transform this, but I did see that it was similar to $L^{-1}[\frac{2a^3}{(s^2+a^2)^2}] = sin(at)-atcos(at)$ which might be helpful
So in the end, how do I do $L^{-1}[\frac{3}{(s^2+9)^2}]$?
Well, using the table of selected Laplace transforms we can see that:
$$\text{y}_\alpha\left(x\right):=\displaystyle\mathscr{L}_\text{s}^{-1}\left[\frac{2\alpha^3}{\left(\text{s}^2+\alpha^2\right)^2}\right]_{\left(x\right)}=\sin\left(\alpha x\right)-\alpha x\cos\left(\alpha x\right)\tag1$$
Now, in your case we have:
$$\frac{3}{\left(\text{s}^2+9\right)^2}=\frac{1}{18}\cdot\frac{2\cdot3^3}{\left(\text{s}^2+3^2\right)^2}\tag2$$