<s>[INST] Use Laplace Transform to solve the following IVP: I know that this is a somewhat simple problem but I have been having trouble coming up with the little "tricks" that help with Laplace. 
The problem is:
$y''+2y' +5y = e^{-t}\sin(2t)$ where $y(0) = 2, y'(0) = -1$
Attempt at Solution
$(s^2+2s+5)Y = \frac{2}{(s+1)^2+4} + (s+2)(2) + 1(-1)$
$Y = \frac{2}{(s^2+2s+5)^2}+\frac{2s+3}{s2+2s+5}$
$Y = \frac{2}{(s^2+2s+5)^2}+2(\frac{s+1}{s^2+2s+5})+\frac{1}{s^2+2s+5}$
$Y = \frac{2}{(s^2+2s+5)^2}+\frac{1}{2}\sin(2t)+2e^{-t}\cos(2t)$
And I am stuck here. 
The answer given in the book is as follows:
Y = $\frac{5}{8}e^{-t}\sin(2t)+2e^{-t}\cos(2t)-\frac{1}{4}te^{-t}\cos(2t)$
Any help would be greatly appreciated. 
 [/INST] We can take inverse Laplace transform by using the Bromwich integral.
That is,
\begin{align}
\mathcal{L}^{-1}\{Y(s)\} &= \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma -iT}^{\gamma +iT}Y(s)e^{st}ds\\
&= \sum\text{Res}
\end{align}
where 
$$
Y(s) = \frac{2}{(s^2 + 2s + 5)^2} + \frac{2s + 3}{s^2 + 2s + 5}
$$
Then the poles of $s$ are at $s = -1\pm 2i$. We can then evaluate the integrals by  splitting them into two. In the first one, the poles are of order two.
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