Question

# find the inverse of Laplace transform frac{3}{(s+2)^2}-frac{2s+6}{(s^2+4)}

Laplace transform
find the inverse of Laplace transform
$$\frac{3}{(s+2)^2}-\frac{2s+6}{(s^2+4)}$$

2021-01-09
Step 1
Given Laplace transform is,
$$\frac{3}{(s+2)^2}-\frac{2s+6}{(s^2+4)}$$
Step 2
Taking inverse of Laplace transformation as,
$$L^{-1}\left\{\frac{3}{(s+2)^2}-\frac{2s+6}{(s^2+4)}\right\}=L^{-1}\left\{\frac{3}{(s+2)^2}\right\}-L^{-1}\left\{\frac{2s+6}{(s^2+4)}\right\}$$
$$=3L^{-1}\left\{\frac{1}{(s+2)^2}\right\}-2L^{-1}\left\{\frac{s}{s^2+4}\right\}-6L^{-1}\left\{\frac{1}{s^2+4}\right\}$$
Step 3
Some formulas of inverse of Laplace transformation are
1) $$L^{-1}\left\{F(s)\right\}=f(t) \text{ then } L^{-1}\left\{F(s-a)\right\}=e^{at}f(t)$$
$$\Rightarrow L^{-1}\left\{\frac{1}{(s+2)^2}\right\}=e^{-2t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{-2t}t$$
2)$$L^{-1}\left\{\frac{a}{s^2+a^2}\right\}=\sin at$$
$$\Rightarrow L^{-1}\left\{\frac{1}{s^2+4}\right\}=\frac{1}{2}L^{-1}\left\{\frac{2}{s^2+4}\right\}=\frac{(\sin2t)}{2}$$
3)$$L^{-1}\left\{sF(s)\right\}=f'(t)+f(0)$$
Then $$L^{-1}\left\{\frac{s}{s^2+4}\right\}=\frac{1}{2}L^{-1}\left\{s\frac{2}{s^2+4}\right\}=\frac{1}{2}((\sin2t)'+\sin(0))=\cos2t$$
Step 4
Then, consider equation (1),
$$3L^{-1}\left\{\frac{1}{(s+2)^2}\right\}-2L^{-1}\left\{\frac{s}{s^2+4}\right\}-6L^{-1}\left\{\frac{1}{s^2+4}\right\}=3L^{-1}\left\{\frac{1}{(s+2)^2}\right\}-2L^{-1}\left\{\frac{s}{s^2+4}\right\}-\frac{6}{2}L^{-1}\left\{\frac{2}{s^2+4}\right\}=3e^{-2t}t-2\cos2t-3\sin2t$$
Step 5
Therefore, the inverse of the given Laplace transformation is,
$$L^{-1}\left\{\frac{3}{(s+2)^2}-\frac{2s+6}{s^2+4}\right\}=3e^{-2t}t-2\cos2t-3\sin2t$$