<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] You need the inverse Laplace transform of 
$$ \dfrac{2}{(s^2 + 2 s + 5)^2}$$
This may help:
$$ {\mathcal L} \{t e^{at} \cos(bt)\} =  \dfrac{(s-a)^2 - b^2}{((s-a)^2+b^2)^2}
= \dfrac{1}{(s-a)^2 + b^2} - \dfrac{2b^2}{((s-a)^2+b^2)^2}$$
</s>