<s>[INST] Laplace transform of a differential equation: $y'+y=e^{-3t}\cos(2t)$ Given $y(0)=0$, solve

$$y'+y=e^{3t}cos(2t)  $$

Steps:
$$sY(s)-y(0)+Y(s)=\frac{s+3}{(s+3)^2+2^2}$$
$$Y(s)(s+1)=\frac{s+3}{(s+3)^2+2^2}$$
$$Y(s)=\frac{s+1}{(s+1)((s+3)^2+2^2)}+\frac{2}{(s+1)((s+3)^2+2^2)}$$
$$Y(s)=\frac{1}{(s+3)^2+2^2}+\frac{\textbf{2}}{\textbf{(s+1)((s+3)^2+2^2)}}$$
Now I can reverse transform the first fraction, may I know how to deal with bolded fraction?
 [/INST] Hint. You may use a partial fraction decomposition to get
$$
\frac{2}{(s+1)((s+3)^2+2^2)}=\frac{1}{4(s+1)}-\frac14\cdot\frac{(s+3)}{(s+3)^2+2^2}-\frac12\cdot\frac{1}{(s+3)^2+2^2}
$$ then each term can be classically reversed.
</s>