Solve initial value problem with unspecified right-hand side $g(t)$ Consider the initial value problem 
$$y''-6y'+9y=g(t),\quad y(0)=1,\ y'(0)=3.$$ 
1) Use the Convolution Theorem to find the solution to the IVP for any piecewise continuous function $g(t)$ that is of exponential order.
2) Using the solution from above, what is the solution of $g(t)=e^{3t}$?
Transform:
$s^2Y(s)-sy(0)-y'(0)-6[sY(s)-y(0)]+9Y(s)=G(s)$
$s^2Y(s)-s-3-6sY(s)+6+9Y(s)=G(s)$
$(s^2-6s+9)Y(s)=s+3+G(s)$
$Y(s)=\frac{s+3}{s^2-6s+9}+\frac{G(s)}{s^2-6s+9} $
I am stuck here. Any help?

You have a $-s-3+6=-s+3$ on the left-hand side, so when it goes to the right-hand side, you get 
$$
Y(s)=\frac{\color{blue}{s-3}}{s^2-6s+9}+\frac{G(s)}{s^2-6s+9}.
$$
Then, since $s^2-6s+9=(s-3)^2$,
\begin{align}
Y(s)&=\frac{s-3}{s^2-6s+9}+\frac{G(s)}{s^2-6s+9}\\
&=\frac{s-3}{(s-3)^2}+\frac{G(s)}{(s-3)^2}\\
&=\frac{1}{s-3}+\frac{G(s)}{(s-3)^2}.\\
\end{align}
Take the inverse Laplace transform of each piece, using the Convolution Theorem on the last term:
\begin{align}
y(t)&=e^{3t}+\mathscr{L^{-1}}\left\{G(s)\cdot\frac{1}{(s-3)^2}\right\}\\
&=e^{3t}+g(t)*te^{3t}\\
&=e^{3t}+\int_0^t g(t-s)se^{3s}\,ds.
\end{align}
When $g(t)=e^{3t}$ (which is what I suspect you meant), this becomes
\begin{align}
y(t)&=e^{3t}+\int_0^t g(t-s)se^{3s}\,ds\\
&=e^{3t}+\int_0^t e^{3(t-s)}se^{3s}\,ds\\
&=e^{3t}+\int_0^t e^{3t}e^{-3s}se^{3s}\,ds\\
&=e^{3t}+e^{3t}\int_0^t s\,ds\\
&=e^{3t}+{1\over 2}t^2e^{3t}.
\end{align}
