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.

\(\int_0^t e^{5(t-u)} \frac{u^2}{2} du = \frac{1}{2} e^{5t} \int_0^t(e^{-5u} \cdot u^2)du\)

\(= \frac{1}{2}e^{5t}\left[-\frac{1}{125}(25t^2e^{-5t}-2(-5te^{-5t}-e^{-5t})-2)\right]\)

\(= -\frac{1}{250}e^{5t}\left[25t^2e^{-5t}+10te^{-5t}+2e^{-5t}-2\right]\)

\(\int_0^t e^{5(t-u)} \frac{u^2}{2} du = L^{-1}(\frac{1}{s^3(s-5)})=-\frac{1}{10}t^2-\frac{1}{25}t-\frac{1}{125}+\frac{1}{125}e^{5t}\)

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.

\(\int_0^t e^{5(t-u)} \frac{u^2}{2} du = \frac{1}{2} e^{5t} \int_0^t(e^{-5u} \cdot u^2)du\)

\(= \frac{1}{2}e^{5t}\left[-\frac{1}{125}(25t^2e^{-5t}-2(-5te^{-5t}-e^{-5t})-2)\right]\)

\(= -\frac{1}{250}e^{5t}\left[25t^2e^{-5t}+10te^{-5t}+2e^{-5t}-2\right]\)

\(\int_0^t e^{5(t-u)} \frac{u^2}{2} du = L^{-1}(\frac{1}{s^3(s-5)})=-\frac{1}{10}t^2-\frac{1}{25}t-\frac{1}{125}+\frac{1}{125}e^{5t}\)