Question

25. In Problems 1-30 use appropriate algebra and Theorem 7.2.1 to find the given inverse Laplace transform.

${\mathcal{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{s^3+5s}\right\}$

Short answer

$\frac{1}{5}-\frac{1}{5}\cos \left(\sqrt{5}t\right)$

Step-by-step solution

Definition of inverse Transform

THEOREM 7.2.1 Some Inverse Transforms

1. $1={\mathcal{L}}^{-1}\left\{\frac{1}{s}\right\}$
   
2. ${\mathrm{t}}^{\mathrm{n}}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{n}!}{{\mathrm{s}}^{\mathrm{n}+1}}\right\},\mathrm{n}=1,2,3,\dots$
   
3. ${\mathrm{e}}^{\mathrm{at}}={\mathcal{L}}^{-1}\left\{\frac{1}{\mathrm{s}-\mathrm{a}}\right\}$
   
4. $\mathrm{sinkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
5. $\mathrm{coskt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+{\mathrm{k}}^2}\right\}$
   
6. $\mathrm{sinhkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{k}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   
7. $\mathrm{coshkt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{k}}^2}\right\}$
   

Use Inverse Transform

${\mathrm{L}}^{-1}\left\{\frac{1}{{\mathrm{s}}^3+5\mathrm{s}}\right\}={\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}\left({\mathrm{s}}^2+5\right)}\right\}$

Here we have to decompose the fraction.

Decomposing

$\frac{1}{\mathrm{s}\left({\mathrm{s}}^2+5\right)}=\frac{\mathrm{a}}{\mathrm{s}}+\frac{\mathrm{bs}+\mathrm{c}}{{\mathrm{s}}^2+5}\to \frac{1}{\mathrm{s}\left({\mathrm{s}}^2+5\right)}=\frac{1}{5\mathrm{s}}-\frac{\mathrm{s}}{5\left({\mathrm{s}}^2+5\right)}$

The trick is to algebraically manipulate the expression to make it look like something we know

Now,

$\frac{1}{5}{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}}\right\}-\frac{1}{5}{\mathrm{L}}^{-1}\left\{\frac{\mathrm{s}}{{\mathrm{s}}^2+5}\right\}=\frac{1}{5}-\frac{1}{5}\cos \left(\sqrt{5}\mathrm{t}\right)$.