Which rule should one use on these two inverse Laplace transforms? I have an ode $y''-6y'+9y=t^2e^{3t}$, which I solved using the Laplace transform. However, when I come to the end result, which I need to do an inverse transform on, I get something which does not appear to be readily compatible with inverse laplace transform tables.
With I.C. $y(0)=2$, $y'(0)=6$:
$\mathscr{L}^{-1}\{y''\}=s^2Y-y'(0)-y(0)=s^2Y-8$
$\mathscr{L}^{-1}\{y'\}=sY-y(0)=sY-2$
$\mathscr{L}^{-1}\{y\}=Y$
This gives:
$s^2Y-6sY+9Y=t^2e^{3t}-4$
Then I get:
\begin{equation}
Y=\frac{t^2e^{3t}}{s^2-6s+9}-4\frac{1}{s^2-6s+9}
\end{equation}
which gives the following inverse:
\begin{equation}
y(t)=\mathscr{L}^{-1}\{\frac{t^2e^{3t}}{(s-3)^2}\}-4\mathscr{L}^{-1}\{\frac{1}{(s-3)^2}\}
\end{equation}
Using wolfram, this gives $y(t)=e^{6t}t^3-4te^{3t}$, but there are some familiarities here that suggest that a specific rule should be applied? If there is, which rule would that be?
Thanks

We have
$$y''-6y'+9y=t^2e^{3t}, y(0) = 2, y'(0) = 6$$
Taking the Laplace Transform
$$(s^2 Y(s) -s y(0) - y'(0)) - 6(s Y(s) -y(0)) + 9 Y(s) = \dfrac{2}{(s-3)^3}$$
Adding the initial conditions
$$(s^2 Y(s) -2s -6) - 6(s Y(s) -2) + 9 Y(s) = \dfrac{2}{(s-3)^3}$$
Grouping terms
$$(s^2 -6s + 9)Y(s) - 2s + 6 = \dfrac{2}{(s-3)^3}$$
Simplifying
$$(s-3)^2 Y(s) = \dfrac{2}{(s-3)^3} + 2 s - 6$$
Solving for $Y(s)$
$$Y(s) = \dfrac{2}{(s-3)^5} + \dfrac{2s}{(s-3)^2} - \dfrac{6}{(s-3)^2}$$
Finding the inverse Laplace transform
$$y(t) = \dfrac{1}{12} e^{3 t} t^4+2 e^{3 t}$$
