find inverse Laplace transform of the given Function

$L}^{-1}\frac{s-5}{{(s-5)}^{2}-9$

Tabansi
2021-09-23
Answered

find inverse Laplace transform of the given Function

$L}^{-1}\frac{s-5}{{(s-5)}^{2}-9$

You can still ask an expert for help

AGRFTr

Answered 2021-09-24
Author has **95** answers

Step 1

We have to find inverse Laplace transform of the given Function

Step 2

We have to find

${L}^{-1}\left[\frac{s-5}{{(s-5)}^{2}-9}\right]$

Solution:${L}^{-1}\left[\frac{s-5}{{(s-5)}^{2}-9}\right]$

By shifting theorem inverse Laplace Transform.

${L}^{-1}\left[F(s-a)\right]={e}^{at}{L}^{-1}\left[F\left(s\right)\right]={e}^{at}f\left(t\right)$

$\therefore {L}^{-1}\left[\frac{s-5}{{(s-5)}^{2}-9}\right]={e}^{st}{L}^{-1}\left[\frac{s}{{s}^{2}-9}\right]\left(1\right)$

$\therefore {L}^{-1}\left[\frac{s}{{s}^{2}-9}\right]={L}^{-1}\left[\frac{s}{{s}^{2}-{3}^{2}}\right]$

$=\mathrm{cos}h\left(3t\right)[\because {L}^{-1}\left[\frac{s}{{s}^{2}-{a}^{2}}\right]=\mathrm{cos}h\left(at\right)]$

use this value in (1)

$\therefore {L}^{-1}\left[\frac{s-5}{{(s-5)}^{2}-9}\right]={e}^{5t}\mathrm{cos}h\left(3t\right)$

We have to find inverse Laplace transform of the given Function

Step 2

We have to find

Solution:

By shifting theorem inverse Laplace Transform.

use this value in (1)

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