# find inverse Laplace transform of the given Function. L^(-1) frac(s-5)((s-5)^2-9)

find inverse Laplace transform of the given Function
${L}^{-1}\frac{s-5}{{\left(s-5\right)}^{2}-9}$
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Step 1
We have to find inverse Laplace transform of the given Function
Step 2
We have to find
${L}^{-1}\left[\frac{s-5}{{\left(s-5\right)}^{2}-9}\right]$
Solution: ${L}^{-1}\left[\frac{s-5}{{\left(s-5\right)}^{2}-9}\right]$
By shifting theorem inverse Laplace Transform.
${L}^{-1}\left[F\left(s-a\right)\right]={e}^{at}{L}^{-1}\left[F\left(s\right)\right]={e}^{at}f\left(t\right)$
$\therefore {L}^{-1}\left[\frac{s-5}{{\left(s-5\right)}^{2}-9}\right]={e}^{st}{L}^{-1}\left[\frac{s}{{s}^{2}-9}\right]\left(1\right)$
$\therefore {L}^{-1}\left[\frac{s}{{s}^{2}-9}\right]={L}^{-1}\left[\frac{s}{{s}^{2}-{3}^{2}}\right]$
$=\mathrm{cos}h\left(3t\right)\left[\because {L}^{-1}\left[\frac{s}{{s}^{2}-{a}^{2}}\right]=\mathrm{cos}h\left(at\right)\right]$
use this value in (1)
$\therefore {L}^{-1}\left[\frac{s-5}{{\left(s-5\right)}^{2}-9}\right]={e}^{5t}\mathrm{cos}h\left(3t\right)$