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

Find the inverse Laplace Tranformation by using convolution theorem for the function $\frac{1}{{s}^{3}\left(s-5\right)}$
You can still ask an expert for help

Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Bella

Step 1
According to the given information, it is required to find the inverse of Laplace transform using convolution theorem.
$\frac{1}{{s}^{3}\left(s-5\right)}$
it is required to find ${L}^{-1}\left(\frac{1}{{s}^{3}\left(s-5\right)}\right)$
Step 2
Solving further:

the inverse laplace of G(s) and H(s) are
${L}^{-1}\left(G\left(s\right)\right)={L}^{-1}\left(\frac{1}{{s}^{3}}\right)=\frac{{t}^{2}}{2}=g\left(t\right)$
${L}^{-1}\left(H\left(s\right)\right)={L}^{-1}\left(\frac{1}{\left(s-5\right)}\right)={e}^{5t}=f\left(t\right)$
Step 3
The convolution theorem states that if G(s) and H(s) are the Laplace transform of g(t) and h(t) then,
$L\left(\left(g\cdot h\right)\left(t\right)\right)=G\left(s\right)H\left(s\right)$
$⇒\left(g\cdot t\right)\left(t\right)={L}^{-1}\left(G\left(s\right)H\left(s\right)\right)$
$⇒g\left(t\right)h\left(t\right)={L}^{-1}\left(F\left(s\right)\right)={\int }_{0}^{t}f\left(t-u\right)g\left(u\right)du$
Step 4
Now solve the integral to find the inverse.
${\int }_{0}^{t}{e}^{5\left(t-u\right)}\frac{{u}^{2}}{2}du=\frac{1}{2}{e}^{5t}{\int }_{0}^{t}\left({e}^{-5u}\cdot {u}^{2}\right)du$
$=\frac{1}{2}{e}^{5t}\left[-\frac{1}{125}\left(25{t}^{2}{e}^{-5t}-2\left(-5t{e}^{-5t}-{e}^{-5t}\right)-2\right)\right]$
$=-\frac{1}{250}{e}^{5t}\left[25{t}^{2}{e}^{-5t}+10t{e}^{-5t}+2{e}^{-5t}-2\right]$
${\int }_{0}^{t}{e}^{5\left(t-u\right)}\frac{{u}^{2}}{2}du={L}^{-1}\left(\frac{1}{{s}^{3}\left(s-5\right)}\right)=-\frac{1}{10}{t}^{2}-\frac{1}{25}t-\frac{1}{125}+\frac{1}{125}{e}^{5t}$