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

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

2021-02-25
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:
$$L^{-1}\left[F(s)\right]=L^{-1}\left[\frac{e^{-2s}}{(s-3)^3}\right]$$
$$\Rightarrow f(t)=L^{-1}\left[\frac{e^{-2s}}{(s-3)^3}\right]$$
Use Inverse Laplace Transform Rule :
If $$L^{-1}\left[F(s)\right]=f(t) , \text{ then } L^{-1}\left[e^{-as}F(s)\right]=H(t-a)f(t-a)$$ 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 $$L^{-1}[F(s)]=f(t) , \text{ then } L^{-1}[F(s-a)]=e^{at}f(t)$$
$$\Rightarrow L^{-1}\left[\frac{1}{(s-3)^3}\right]=e^{3t}L^{-1}\left[\frac{1}{s^3}\right]=e^{3t}L^{-1}\left[\frac{2}{2s^3}\right]$$
$$\Rightarrow L^{-1}\left[\frac{1}{(s-3)^3}\right]=e^{3t}\frac{1}{2}L^{-1}\left[\frac{2}{s^3}\right]$$
$$\Rightarrow L^{-1}[\frac{1}{(s-3)^3}]=e^{3t}\frac{1}{2t^2} \left\{\text{ Using } L^{-1}\left[\frac{n!}{s^(n+1)}\right]=t^n \right\}$$
$$\Rightarrow L^{-1}\left[\frac{1}{(s-3)^3}\right]=\frac{t^2}{2e^{3t}}$$
Hence, we get:
$$f(t)=H(t-2)\left(\frac{t-2}{2}\right)e^{3(t-2)}$$
Step 3
Inverse Laplace Transform is given by:
$$f(t)=H(t-2)\left(\frac{t-2}{2}\right)e^{3(t-2)}$$

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Ans. $$F(s)=\left(\frac{1}{s^2}+\frac{2}{s}\right)e^{-2s}$$