Solving an IVP with Laplace Transforms I'm trying to solve the following IVP (differential equations) with the Laplace Transform method:
\begin{cases}
    y''+9y=36t\sin(3t)\\
    y(0) = 0\\
    y'(0) = 3
\end{cases}
After taking the Laplace Transform of both sides, I obtain
$$s^2Y(s) - 3 + 9sY(s) = \frac{216}{(s^2+9)^2}$$
where $Y(s) = \mathcal{L}[y(t)]$.
Solving for $Y(s)$, I get the following equation:
$$Y(s) = \frac{216}{(s^2+9)^2(s^2+9s)} + \frac{3}{s^2+9s}$$
When I try to take the inverse Laplace transform of both sides to solve for $y(t)$, I can't figure out what to do with the first term on the RHS (second term is a simple partial fraction decomposition). The algorithm for partial fractions with linear terms is problematic because the numerator of the first term has a $\frac{1}{0}$ issue. Without partial fractions, the only way I can think to do this one is through Reduction of Order, but my book only gives examples of the case where we only have $(s^2+b^2)^{k+1}$ in the denominator. The thing that's messing me up so much is the $s^2+9s$ term. Any help or insight in how to solve this problem from here would be much appreciated.
EDIT: After probably 2 hours of mulling over this problem, I realized that I had forgotten the $s$ after 216 because of the chain rule when taking the derivative of transforms. That makes this so much easier. I think I can do it.
 A: Notice:


*

*$$\mathcal{L}_{t}\left[y'(t)\right]_{(s)}=sy(s)-y(0)$$

*$$\mathcal{L}_{t}\left[y''(t)\right]_{(s)}=s^2y(s)-sy(0)-y'(0)$$

*$$\mathcal{L}_{t}\left[t\right]_{(s)}=\frac{1}{s^2}$$

*$$\mathcal{L}_{t}\left[\sin(t)\right]_{(s)}=\frac{1}{1+s^2}$$

*$$\mathcal{L}_{t}\left[\sin(at)\right]_{(s)}=\frac{a}{a^2+s^2}$$

*$$\mathcal{L}_{t}\left[t\sin(at)\right]_{(s)}=\frac{2as}{\left(a^2+s^2\right)^2}$$



Solving your question:
$$y''(t)+9y(t)=36t\sin(3t)\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[y''(t)+9y(t)\right]_{(s)}=\mathcal{L}_{t}\left[36t\sin(3t)\right]_{(s)}\Longleftrightarrow$$
$$\mathcal{L}_{t}\left[y''(t)\right]_{(s)}+9\mathcal{L}_{t}\left[y(t)\right]_{(s)}=36\mathcal{L}_{t}\left[t\sin(3t)\right]_{(s)}\Longleftrightarrow$$
$$s^2y(s)-sy(0)-y'(0)+9y(s)=36\cdot\frac{6s}{\left(9+s^2\right)^2}\Longleftrightarrow$$

Now, set $y(0)=0$ and $y'(0)=3$:  

$$s^2y(s)-s\cdot0-3+9y(s)=\frac{216s}{\left(9+s^2\right)^2}\Longleftrightarrow$$
$$s^2y(s)-3+9y(s)=\frac{216s}{\left(9+s^2\right)^2}\Longleftrightarrow$$
$$s^2y(s)+9y(s)=\frac{216s}{\left(9+s^2\right)^2}+3\Longleftrightarrow$$
$$y(s)\left[s^2+9\right]=\frac{216s}{\left(9+s^2\right)^2}+3\Longleftrightarrow$$
$$y(s)=\frac{\frac{216s}{\left(9+s^2\right)^2}+3}{s^2+9}\Longleftrightarrow$$
$$y(s)=\frac{3\left(72s+\left(9+s^2\right)^2\right)}{\left(9+s^2\right)^3}\Longleftrightarrow$$
$$\mathcal{L}_{s}^{-1}\left[y(s)\right]_{(t)}=\mathcal{L}_{s}^{-1}\left[\frac{3\left(72s+\left(9+s^2\right)^2\right)}{\left(9+s^2\right)^3}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=3\mathcal{L}_{s}^{-1}\left[\frac{72s+\left(9+s^2\right)^2}{\left(9+s^2\right)^3}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=3\mathcal{L}_{s}^{-1}\left[\frac{72s}{\left(9+s^2\right)^3}+\frac{1}{s^2+9}\right]_{(t)}\Longleftrightarrow$$
$$y(t)=3\left[72\mathcal{L}_{s}^{-1}\left[\frac{s}{\left(9+s^2\right)^3}\right]_{(t)}+\mathcal{L}_{s}^{-1}\left[\frac{1}{s^2+9}\right]_{(t)}\right]\Longleftrightarrow$$
$$y(t)=3\left[\frac{t\left(\sin(3t)-3t\cos(3t)\right)}{3}+\frac{\sin(3t)}{3}\right]\Longleftrightarrow$$
$$y(t)=t\left(\sin(3t)-3t\cos(3t)+\sin(3t)\right)\Longleftrightarrow$$
$$y(t)=(1+t)\sin(3t)-3t^2\cos(3t)$$
