Laplace transform of a differential equation: $y'+y=e^{-3t}\cos(2t)$

Question

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?

Answer 1

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.