Find the inverse Laplace Tranformation by using convolution theorem for the function frac{1}{s^3(s-5)}

Question
Laplace transform
asked 2021-03-05
Find the inverse Laplace Tranformation by using convolution theorem for the function \(\frac{1}{s^3(s-5)}\)

Answers (1)

2021-03-06
Step 1
According to the given information, it is required to find the inverse of Laplace transform using convolution theorem.
\(\frac{1}{s^3(s-5)}\)
it is required to find \(L^{-1}\left(\frac{1}{s^3(s-5)}\right)\)
Step 2
Solving further:
\(\text{let } G(s)=\frac{1}{s^3} , H(s)=\frac{1}{(s-5)}\)
the inverse laplace of G(s) and H(s) are
\(L^{-1}(G(s))=L^{-1}\left(\frac{1}{s^3}\right)=\frac{t^2}{2}=g(t)\)
\(L^{-1}(H(s))=L^{-1}\left(\frac{1}{(s-5)}\right)=e^{5t}=f(t)\)
Step 3
The convolution theorem states that if G(s) and H(s) are the Laplace transform of g(t) and h(t) then,
\(L((g*h)(t))=G(s)H(s)\)
\(\Rightarrow (g*t)(t)=L^{-1}(G(s)H(s))\)
\(\Rightarrow g(t)h(t)=L^{-1}(F(s))=\int_0^t f(t-u)g(u)du\)
Step 4
Now solve the integral to find the inverse.
\(\int_0^t e^{5(t-u)} \frac{u^2}{2} du = \frac{1}{2} e^{5t} \int_0^t(e^{-5u} \cdot u^2)du\)
\(= \frac{1}{2}e^{5t}\left[-\frac{1}{125}(25t^2e^{-5t}-2(-5te^{-5t}-e^{-5t})-2)\right]\)
\(= -\frac{1}{250}e^{5t}\left[25t^2e^{-5t}+10te^{-5t}+2e^{-5t}-2\right]\)
\(\int_0^t e^{5(t-u)} \frac{u^2}{2} du = L^{-1}(\frac{1}{s^3(s-5)})=-\frac{1}{10}t^2-\frac{1}{25}t-\frac{1}{125}+\frac{1}{125}e^{5t}\)
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