Question

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

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

Short answer

$\frac{1+\sqrt{3}}{2}{e}^{-\sqrt{3}t}+\frac{1-\sqrt{3}}{2}{e}^{\sqrt{3}t}$

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. $t^n={\mathcal{L}}^{-1}\left\{\frac{n!}{s^{n+1}}\right\},n=1,2,3,\dots$
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

$L^{-1}\left\{\frac{s-3}{(s-\sqrt{3})(s+\sqrt{3})}\right\}$

Here we have to decompose the fraction.

Decomposing

$\frac{\mathrm{s}-3}{(\mathrm{s}-\sqrt{3})(\mathrm{s}+\sqrt{3})}=\frac{\mathrm{a}}{\mathrm{s}-\sqrt{3}}+\frac{\mathrm{b}}{\mathrm{s}+\sqrt{3}}\to \frac{\mathrm{s}-3}{(\mathrm{s}-\sqrt{3})(\mathrm{s}+\sqrt{3})}=\frac{3+\sqrt{3}}{2\sqrt{3}(\mathrm{s}+\sqrt{3})}-\frac{3-\sqrt{3}}{2\sqrt{3}(\mathrm{s}-\sqrt{3})}$

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

Now, $\frac{3}{2\sqrt{3}}{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-(-\sqrt{3})}\right\}+\frac{1}{2}{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-(-\sqrt{3})}\right\}-\frac{3}{2\sqrt{3}}{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-\sqrt{3}}\right\}+\frac{1}{2}{\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-\sqrt{3}}\right\}$

$=\frac{\sqrt{3}}{2}{\mathrm{e}}^{-\sqrt{3}\mathrm{t}}+\frac{1}{2}{\mathrm{e}}^{-\sqrt{3}\mathrm{t}}-\frac{\sqrt{3}}{2}{\mathrm{e}}^{\sqrt{3}\mathrm{t}}+\frac{1}{2}{\mathrm{e}}^{\sqrt{3}\mathrm{t}}$

$=\frac{1+\sqrt{3}}{2}{\mathrm{e}}^{-\sqrt{3}\mathrm{t}}+\frac{1-\sqrt{3}}{2}{\mathrm{e}}^{\sqrt{3}\mathrm{t}}$