<s>[INST] Use Laplace Transformations to solve $y''+2y'+5y=3e^{-x}sin(x)$, with $y(0)=0$, $y'(0)=3$ I've gotten this far and I cannot proceed:
$L[y]=\frac{L[3e^{-x}sin(x)]+3}{p^2+2p+5}= \frac{3}{((p+1)^2+1)(p^2+2p+5)}+\frac{3}{p^2+2p+5}$
I'm finding it impossible to find the inverse to solve for $y$. I would guess partial fraction decomposition is next, but I cannot do it for these quantities. 
Help?
 [/INST] It's easier than it looks. Just write out
$$\frac3{(p^p+2p+5)(p^2+2p+2)}=\frac{Ap+B}{p^2+2p+5}+\frac{Cp+D}{p^2+2p+2}$$
Multiply out that denominator
$$3=(Ap+B)(p^2+2p+2)+(Cp+D)(p^2+2p+5))$$
Then compare coefficients of like powers of $p$
$$\begin{align}0 & =A+C&(1)\\
0 &=2A+B+2C+D&(2)\\
0 &=2A+2B+5C+2D&(3)\\
3 &=2B+5D&(4)\end{align}$$
Now subtract twice eq $(1)$ from eqs $(2)$ and $(5)$
$$\begin{align}0 &=B+D&(5)\\
0&=2B+3C+2D&(6)\end{align}$$
Subtract twice eq $(5)$ from eqs $(6)$ and $(4)$
$$\begin{align}0&=3C&(7)\\
3&=3D&(8)\end{align}$$
And back into eq $(1)$ we have
$$\begin{align}0&=A&(9)\end{align}$$
At this point from eq (8), we can see that $D=1$ and then from eq $(5)$, $B=-1$. So
$$\frac3{(p^p+2p+5)(p^2+2p+2)}=\frac{-1}{p^2+2p+5}+\frac{1}{p^2+2p+2}$$
So I hope you can continue from this point.
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