# Solve for the Inverse Laplace Transformations. Show the solution. F(s)=frac{6}{(s^2+4s+20)^2}

Solve for the Inverse Laplace Transformations. Show the solution.
$F\left(s\right)=\frac{6}{\left({s}^{2}+4s+20{\right)}^{2}}$
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Step 1
Given,
$F\left(s\right)=\frac{6}{\left({s}^{2}+4s+20{\right)}^{2}}$
Find the inverse Laplace Transform of the given function.
Step 2
Now
$F\left(s\right)=\frac{6}{\left({s}^{2}+4s+20{\right)}^{2}}$
$=\frac{6}{\left(\left({s}^{2}+4s+4\right)+16{\right)}^{2}}$
$=\frac{6}{\left(\left(s+2{\right)}^{2}+{4}^{2}{\right)}^{2}}$
Taking inverse Laplace Transform of both sides,
${L}^{-1}\left[F\left(s\right)\right]={L}^{-1}\left[\frac{6}{\left(\left(s+2{\right)}^{2}+{4}^{2}{\right)}^{2}}\right]$
$f\left(t\right)={L}^{-1}\left[\frac{6}{\left(\left(s+2{\right)}^{2}+{4}^{2}{\right)}^{2}}\right]$
Step 3 Use the formula such that
${L}^{-1}\left[F\left(s\right)\right]=f\left(t\right)$
$⇒{L}^{-1}\left[F\left(s-a\right)\right]={e}^{at}f\left(t\right)$
Then,
$f\left(t\right)={L}^{-1}\left[\frac{6}{\left(\left(s+2{\right)}^{2}+{4}^{2}{\right)}^{2}}\right]$
$={e}^{-2t}{L}^{-1}\left[\frac{6}{\left({s}^{2}+{4}^{2}{\right)}^{2}}\right]$
Again the formula,
${L}^{-1}\left[\frac{2{a}^{3}}{\left({s}^{2}+{a}^{2}{\right)}^{2}}\right]=\mathrm{sin}\left(at\right)-at\mathrm{cos}\left(at\right)$
Step 4
So,
$f\left(t\right)={e}^{-2t}{L}^{-1}\left[\frac{6}{\left({s}^{2}+{4}^{2}{\right)}^{2}}\right]$
$=\frac{3}{{4}^{3}}{e}^{-2t}{L}^{-1}\left[\frac{2\cdot {4}^{3}}{\left({s}^{2}+{4}^{2}{\right)}^{2}}\right]$
$=\frac{3}{64}{e}^{-2t}\left[\mathrm{sin}\left(4t\right)-4t\mathrm{cos}\left(4t\right)\right]$
Hence, $=\frac{3}{64}{e}^{-2t}\left[\mathrm{sin}\left(4t\right)-4t\mathrm{cos}\left(4t\right)\right]$