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

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