2022-09-06

Inverse Laplace Transform partial fraction ω2(s2+ω2)(s2+ω2)

Shania Delacruz

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

Kathy Guerra

Hint:
Use $\frac{{\omega }^{2}}{\left({s}^{2}+{\omega }^{2}{\right)}^{2}}=\frac{1}{2}\left[\frac{{s}^{2}+{\omega }^{2}}{\left({s}^{2}+{\omega }^{2}{\right)}^{2}}-\frac{{s}^{2}-{\omega }^{2}}{\left({s}^{2}+{\omega }^{2}{\right)}^{2}}\right]=\frac{1}{2}\left[\frac{1}{{s}^{2}+{\omega }^{2}}+\frac{d}{ds}\left(\frac{s}{{s}^{2}+{\omega }^{2}}\right)\right]$
Now use ${\mathcal{L}}^{-1}\left(\frac{\omega }{{s}^{2}+{\omega }^{2}}\right)=\mathrm{sin}\left(\omega t\right)$ and ${\mathcal{L}}^{-1}\left(\frac{s}{{s}^{2}+{\omega }^{2}}\right)=\mathrm{cos}\left(\omega t\right)$

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