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Can anybody help me with the answer of this question?

Find the inverse Laplace transform of $\frac {1}{(s-3)^4}$

4 Answers 4

3

This will surely help:

Let $F(t)=\mathscr L^{-1}\{f(s)\}$. Then $\mathscr L^{-1}\{f(s-a)\}=e^{at} F(t)$

2

A related problem. The inverse Laplace transform is defined by,

$ f(x) = \frac{1}{2\pi i}\int_C F(s) {\rm e}^{sx} ds \,$

where $F(s)$ is the Laplace transform of $f(x)$, and $C$ is the Bromwich contour (see below).

$ f(x) = \frac{1}{2\pi i} \int_C \frac{{\rm e}^{sx}}{(s-2)^4} ds = \frac{1}{3!} \frac{d^3 }{ds^3} {\rm e}^{sx} |_{s=2} = \frac{1}{6} x^3 {\rm e}^{2x}\,,$

since the integral has a pole of order $4$ at $s=2$, and using the residue theorem.

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1

Take advantage of formal properties of the Laplace transform:

  • If $\mathcal{L}_s(f(x)) = g(s)$, then $\mathcal{L}_s(\mathrm{e}^{a x} f(x)) = g(s-a)$

Also use the table entry:

  • $\mathcal{L}_s(x^n) = (n-1)! s^{-n}$
0

$ \mathcal{L}[t^n]= \frac{\Gamma(n+1)}{s^{n+1}} $ if n is natural number we can write : $ \mathcal{L}[t^n]= \frac{n!}{s^{n+1}} $ and we know: $ \mathcal{L}[e^{at}f(t)]=F(s-a) $ therefor: $ \frac {1}{(s-3)^4} = \frac { \frac{3!}{3!} }{(s-3)^{3+1}} =\frac{1}{3!}\frac{3!}{(s-3)^{3+1}}$ $ \mathcal{L^{-1}} \left[ \frac{1}{3!}\frac{3!}{(s-3)^{3+1}} \right] = \frac{1}{3!} t^3 e^{3t} $