<s>[INST] Laplace transformation $y''+2y'+2y=3\sin x+\cos x$ Given$$y''+2y'+2y=3\sin x+\cos x$$
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$$Y(s)(s^2+2s+2)=\frac{3}{s^2+1}+\frac{s}{s^2+1}-s-2$$
$$Y(s)((s^2+2s+1)+1)=\frac{3}{s^2+1}+\frac{s}{s^2+1}-s-2$$
$$Y(s)((s+1)^2+1)=\frac{3}{s^2+1}+\frac{s}{s^2+1}-s-2$$
$$Y(s)=\frac{3}{(s^2+1)((s+1)^2+1)}+\frac{s}{(s^2+1)((s+1)^2+1)}-\frac{s}{((s+1)^2+1)}-\frac{2}{((s+1)^2+1)}$$
Now I need the inverse of $$\frac{1}{((s+1)^2+1)}$$ to continue. Can't find it anywhere
 [/INST] Hint: $\mathscr{L^{-1}}\left(\frac{b}{(s-a)^2+b^2}\right) = e^{at}\sin{(bt)}$; in your case, $a=-1$ and $b=1$. You can prove it by the integral definition of Laplace transformation. More info here.
Another hint:, you can also find the inverse of $\frac{s}{((s+1)^2+1)}$ by smart usage of the $\mathscr{L^{-1}}$ of $e^{at}\cos{(bt)}$ in the link above. (That is, $s=s+0=s+(1-1)$)
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