# determine the inverse Laplace transform of F. F(s)=frac{e^{-2s}}{(s-3)^3}

determine the inverse Laplace transform of F.
$F\left(s\right)=\frac{{e}^{-2s}}{\left(s-3{\right)}^{3}}$
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Step 1
Let f(t) be the inverse Laplace transform of the given function
$F\left(s\right)=\frac{{e}^{-2s}}{\left(s-3{\right)}^{3}}$
Take inverse Laplace transform on both sides:
${L}^{-1}\left[F\left(s\right)\right]={L}^{-1}\left[\frac{{e}^{-2s}}{\left(s-3{\right)}^{3}}\right]$
$⇒f\left(t\right)={L}^{-1}\left[\frac{{e}^{-2s}}{\left(s-3{\right)}^{3}}\right]$
Use Inverse Laplace Transform Rule :
If where , H(t) is Heavside step function
Now, we have a = 2 in our case
Hence,
$f\left(t\right)=H\left(t-2\right){L}^{-1}\left[\frac{1}{\left(s-3{\right)}^{3}}\right]$
Step 2
Calculation of inverse Laplace Transform of function on right side:
Apply inverse Laplace Transform rule : If
$⇒{L}^{-1}\left[\frac{1}{\left(s-3{\right)}^{3}}\right]={e}^{3t}{L}^{-1}\left[\frac{1}{{s}^{3}}\right]={e}^{3t}{L}^{-1}\left[\frac{2}{2{s}^{3}}\right]$
$⇒{L}^{-1}\left[\frac{1}{\left(s-3{\right)}^{3}}\right]={e}^{3t}\frac{1}{2}{L}^{-1}\left[\frac{2}{{s}^{3}}\right]$

$⇒{L}^{-1}\left[\frac{1}{\left(s-3{\right)}^{3}}\right]=\frac{{t}^{2}}{2{e}^{3t}}$
Hence, we get:
$f\left(t\right)=H\left(t-2\right)\left(\frac{t-2}{2}\right){e}^{3\left(t-2\right)}$
Step 3
Inverse Laplace Transform is given by:
$f\left(t\right)=H\left(t-2\right)\left(\frac{t-2}{2}\right){e}^{3\left(t-2\right)}$