Inverse Laplace Transform partial fraction $\frac{\omega ^{2}}{\left ( s^{2}+\omega ^{2} \right )( s^{2}+\omega ^{2} )}$

Question

While solving a second order differential equation, I have reached at a stage where I have to calculate the inverse laplace transform of $\frac{\omega ^{2}}{\left ( s^{2}+\omega ^{2} \right )( s^{2}+\omega ^{2} )}$. Can anyone help me to decompose the given expression into partial fractions so that I can take its inverse laplace transform. Or, is there any other easy methods to find its inverse laplace transform.

Answer 1

Differentiation is a good short-cut for this case: $$ \mathscr{L}\left\{\sin(wt)\right\} = \frac{w}{s^{2}+w^{2}}. $$ Differentiate with respect to $w$: $$ \begin{align} \mathscr{L}\{t\cos(wt)\} = \frac{d}{dw}\frac{w}{s^{2}+w^{2}} &= -\frac{2w^{2}}{(s^{2}+w^{2})^{2}}+\frac{1}{s^{2}+w^{2}} \\ & = -\frac{2w^{2}}{(s^{2}+w^{2})^{2}}+\mathscr{L}\{\cos(wt)\} \end{align} $$ Now you can solve for $w^{2}/(s^{2}+w^{2})^{2}$ as the Laplace transform of something.

Answer 2

Mathematica gives me

$ \dfrac{-(t w \cos(w t)) + \sin(w t)}{2 w} $

Hint: $w/(s^2+w^2)$ is the transform of $sin(wt)$. CAn you now use the convolution theorem?

Answer 3

Hint:

Use $\frac{\omega^2}{(s^2+\omega^2)^2}=\frac{1}{2}\big[\frac{s^2+\omega^2}{(s^2+\omega^2)^2}-\frac{s^2-\omega^2}{(s^2+\omega^2)^2}\big]=\frac{1}{2}\big[\frac{1}{s^2+\omega^2}+\frac{d}{ds}\big(\frac{s}{s^2+\omega^2}\big)\big]$

Now use $\mathscr{L}^{-1}(\frac{\omega}{s^2+\omega^2})=\sin(\omega t)$ and $\mathscr{L}^{-1}(\frac{s}{s^2+\omega^2})=\cos(\omega t)$.