Question

In Problems 31–34 find the given inverse Laplace transform by fnding the Laplace transform of the indicated function f.

$\mathbf{31}\mathbf{.}{\mathbf{L}}^{\mathbf{-}\mathbf{1}}\mathbf{\left\{}\frac{\mathbf{1}}{{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{a}\mathbf{)}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}}\mathbf{\right\}}\mathbf{;}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{ot}}\mathbf{sinhbt}$

Short answer

${\mathcal{L}}^{-1}\left\{\frac{1}{{(\mathrm{s}-\mathrm{a})}^2-{\mathrm{b}}^2}\right\}=\frac{1}{\mathrm{b}}{\mathrm{e}}^{\mathrm{at}}\mathrm{sinhbt}$

Step-by-step solution

To Find use the formula for laplace tranform

Use the formula for Laplace transform:

$$\begin{gathered} \mathcal{L}\left\{\mathrm{f}\right(\mathrm{t}\left)\right\}={\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}\mathrm{f}\left(\mathrm{t}\right)\text{d}\mathrm{t}\left(\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{at}}\mathrm{sinhbt}\right) \\ ={\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}\left({\mathrm{e}}^{\mathrm{at}}\mathrm{sinhbt}\right)\text{d}\mathrm{t}\left(\mathrm{sinhbt}=\frac{{\mathrm{e}}^{\mathrm{bt}}-{\mathrm{e}}^{-\mathrm{bt}}}{2}\right) \\ ={\int }_0^{\infty }{\mathrm{e}}^{-\mathrm{st}}{\mathrm{e}}^{\mathrm{at}}\left(\frac{{\mathrm{e}}^{\mathrm{bt}}-{\mathrm{e}}^{-\mathrm{bt}}}{2}\right)\text{d}\mathrm{t}\left({\mathrm{e}}^{\mathrm{a}+\mathrm{b}}={\mathrm{e}}^{\mathrm{a}}{\mathrm{e}}^{\mathrm{b}}\right) \end{gathered}$$

$=\frac{1}{2}{\int }_0^{\infty }{e}^{-st}\left(e^{t(a+b)}-e^{t(a-b)}\right)\text{d}t$

$=\frac{1}{2}\left[{\int }_0^{\infty }{e}^{-st}{e}^{t(a+b)}\text{d}t-{\int }_0^{\infty }{e}^{-st}{e}^{t(a-b)}\text{d}t\right]$

$=\frac{1}{2}\left[{\int }_0^{\infty }{e}^{-t(s-(a+b\left)\right)}\text{d}t-{\int }_0^{\infty }{e}^{-t(s-(a-b\left)\right)}\text{d}t\right]$

$=\frac{1}{2}\left[-{\left.\frac{1}{\mathrm{s}-(\mathrm{a}+\mathrm{b})}{\mathrm{e}}^{-\mathrm{t}(\mathrm{s}-(\mathrm{a}+\mathrm{b}\left)\right)}\right|}_0^{\infty }+{\left.\frac{1}{\mathrm{s}-(\mathrm{a}-\mathrm{b})}{\mathrm{e}}^{-\mathrm{t}(\mathrm{s}-(\mathrm{a}-\mathrm{b}\left)\right)}\right|}_0^{\infty }\right]$

$=\frac{1}{2}\left[-\frac{1}{\mathrm{s}-(\mathrm{a}+\mathrm{b})}\left({\mathrm{e}}^{-\infty }-{\mathrm{e}}^0\right)+\frac{1}{\mathrm{s}-(\mathrm{a}-\mathrm{b})}\left({\mathrm{e}}^{-\infty }-{\mathrm{e}}^0\right)\right]\left({\mathrm{e}}^{-\infty }=0,{\mathrm{e}}^0=1\right)$

$=\frac{1}{2}\left[\frac{1}{\mathrm{s}-(\mathrm{a}+\mathrm{b})}-\frac{1}{\mathrm{s}-(\mathrm{a}-\mathrm{b})}\right]$

$=\frac{1}{2}\left[\frac{1}{(\mathrm{s}-\mathrm{a})-\mathrm{b}}-\frac{1}{(\mathrm{s}-\mathrm{a})+\mathrm{b}}\right]$

localid="1664029659542" $=\frac{1}{2}\left[\frac{s-a+\mathrm{b}-s+a+b}{\left(\right(\mathrm{s}-\mathrm{a})-\mathrm{b})\left(\right(\mathrm{s}-\mathrm{a})+\mathrm{b})}\right]$

$=\frac{1}{2}\left[\frac{2\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2-{\mathrm{b}}^2}\right]$

$=\frac{\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2-{\mathrm{b}}^2}$

Divide each side of the equation

Hence,

$\mathcal{L}\left\{{\mathrm{e}}^{\mathrm{at}}\mathrm{sinhbt}\right\}=\frac{\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2-{\mathrm{b}}^2}$

Take inverse Laplace transform of both side.

${\mathrm{e}}^{\mathrm{at}}\mathrm{sinhbt}={\mathcal{L}}^{-1}\left\{\frac{\mathrm{b}}{{(\mathrm{s}-\mathrm{a})}^2-{\mathrm{b}}^2}\right\}$

${\mathrm{e}}^{\mathrm{at}}\mathrm{sinhbt}==\mathrm{b}{\mathcal{L}}^{-1}\left\{\frac{1}{{(\mathrm{s}-\mathrm{a})}^2-{\mathrm{b}}^2}\right\}$

Divide each side of the equation by ,,,$\mathrm{b}$:

${\mathcal{L}}^{-1}\left\{\frac{1}{{(\mathrm{s}-\mathrm{a})}^2-{\mathrm{b}}^2}\right\}=\frac{1}{\mathrm{b}}{\mathrm{e}}^{\mathrm{at}}\mathrm{sinhbt}$

Final proof

${\mathcal{L}}^{-1}\left\{\frac{1}{{(\mathrm{s}-\mathrm{a})}^2-{\mathrm{b}}^2}\right\}=\frac{1}{\mathrm{b}}{\mathrm{e}}^{\mathrm{at}}\mathrm{sinhbt}$