Solve the initial value problem $9y''(t)-2y'(t)+y(t)=t\cdot e^{-t/4}; y(0)=0, y'(0)=1$ using Laplace transformations Solve the initial value problem $9y''(t)-2y'(t)+y(t)=t\cdot e^{-t/4}; y(0)=0, y'(0)=1$ using Laplace transformations
I rearranged for $\bar y(s)=\dfrac{1}{(s+1/4)^2(9s^2-2s+1)}+\dfrac{9}{(9s^2-2s+1)}$
For the first term I solved using partial fractions and obtained:
$\bar f_1(s)=\dfrac{1664}{1089(s+1/4)}+\dfrac{16}{33(s+1/4)^2}-\dfrac{1664s}{121(9s^2-2s+1)}+\dfrac{2380}{1089(9s^2-2s+1)}$
When I tried finding the inverse Laplace of these, the first two terms were fine but when I reached the 3rd and 4th term I felt like the amount of work required was worth too much for what it's worth... have I made a mistake anywhere?
For the second term, I obtained:
$f_2(t)=e^{t/9}\cdot \dfrac{9}{2\sqrt2} \cdot \sin (\dfrac{2\sqrt2}{9})$
edit1: fixed partial fraction decomposition

Noting
$$ \dfrac{s}{9s^2-2s+1}=\frac19\frac{s}{(s-\frac19)^2+(\frac{2\sqrt2}{9})^2},\dfrac{1}{9s^2-2s+1}=\frac19\frac{1}{(s-\frac19)^2+(\frac{2\sqrt2}{9})^2}$$
one has
\begin{eqnarray}
&&L^{-1}(\dfrac{s}{9s^2-2s+1})\\
&=&\frac19L^{-1}(\frac{s}{(s-\frac19)^2+(\frac{2\sqrt2}{9})^2})\\
&=&\frac19L^{-1}(\frac{s-\frac19}{(s-\frac19)^2+(\frac{2\sqrt2}{9})^2})+\frac1{2\sqrt2}L^{-1}(\frac{\frac{2\sqrt2}{9}}{(s-\frac19)^2+(\frac{2\sqrt2}{9})^2})\\
&=&\frac19e^{\frac19t}\cos(\frac{2\sqrt2}{9}t)+\frac1{2\sqrt2}e^{\frac19t}\sin(\frac{2\sqrt2}{9}t)
\end{eqnarray}
and
\begin{eqnarray}
&&L^{-1}(\dfrac{1}{9s^2-2s+1})\\
&=&\frac1{2\sqrt2}L^{-1}(\frac{\frac{2\sqrt2}{9}}{(s-\frac19)^2+(\frac{2\sqrt2}{9})^2})\\
&=&\frac1{2\sqrt2}e^{\frac19t}\sin(\frac{2\sqrt2}{9}t)
\end{eqnarray}
