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

Solution 1:

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

$$y(t) = \dfrac{1}{12} e^{3 t} t^4+2 e^{3 t}$$