Solve $L^{-1}[\frac{1}{p(p^2+4)^2}]$

Question

Solve $L^{-1}[\frac{1}{p(p^2+4)^2}]$

I tried resolving it to partial fractions but could not do. Pls provide me any hint.

Answer 1

Let us make use of three results:

• $$L^{-1}\left[\frac1{s^2+4}\right]=\frac12 \sin 2t$$

• $$L^{-1}\left[\frac1{s(s^2+4)}\right] $$ $$= \int_{0}^{t}\frac1{2} \sin 2t \, dt= \frac14 \left(1-\cos 2t\right)$$

• $$L^{-1} \left[\frac1{s(s^2+4)}\times \frac1{s^2+4} \right] $$ $$= \frac18 \int_{0}^{t} \left(1-\cos 2\tau\right) \sin 2(t-\tau) \, d\tau$$

which gives us the required answer:

$$\frac1{16} \left(1-t\sin 2t - \cos 2t \right) $$