Inverse Laplace Transform of $L^{-1} \Bigl(\frac{1}{s^2 (s+2)^2 }\Bigr)$

Question

How can we find inverse laplace transform of this equation ?

$L^{-1} \Bigl(\frac{1}{s^2 (s+2)^2 }\Bigr)$

Answer 1

Check $$\frac{1}{s^2(s+2)^2}=\frac{1}{4}\left (\frac{1}{s^2}-\frac{1}{s} + \frac{1}{s+2}+\frac{1}{(s+2)^2}\right)$$ use the fact that $$L^{-1} \frac{1}{s}=1,~~ L^{-1} \frac{1}{s^2}=t,~~ L^{-1} \frac{1}{x+2}=e^{-2t}$$ So $$L^{-1}\frac{1}{s^2(s+2)^2}=\frac{1}{4}[t-1+e^{-2t}+te^{-2t}]$$

Answer 2

HINT: Use partial fractions as follws $$\frac{1}{s^2(s+2)^2}=\frac14\left(\frac{1}{s^2}-\frac1s+\frac{1}{s+2}+\frac{1}{(s+2)^2}\right)$$