Question

In Problems 3 - 24 fill in the blanks or answer true or false.

15. ${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{{\left(s-5\right)}^3}\right\}$=……

Short answer

$L^{-1}\left\{\frac{1}{{\left(s-5\right)}^3}\right\}=\frac{e^{5t}{t}^2}{2}$

Step-by-step solution

Consider the function

The function is,

$\begin{array}{l}{L}^{-1}\left\{\frac{1}{{\left(s-5\right)}^3}\right\}\\ \mathrm{For} \frac{1}{{\left(s-5\right)}^3}:a=5, F\left(s\right)=\frac{1}{s^3}\end{array}$

Compute the Laplace inverse

$\begin{array}{l}{L}^{-1}\left\{\frac{1}{{\left(s-5\right)}^3}\right\}\\ \mathrm{If} L^{-1}\left\{F\left(s\right)\right\}=f\left(t\right) \mathrm{then} L^{-1}\left\{F\left(s-a\right)\right\}=e^{at}f\left(t\right)\\ \mathrm{For} \frac{1}{{\left(s-5\right)}^3}:a=5, F\left(s\right)=\frac{1}{s^3}\\ Thus,\end{array}$

$L^{-1}\left\{\frac{1}{{\left(s-5\right)}^3}\right\}=e^{5t}{L}^{-1}\left\{\frac{1}{s^3}\right\}$

subsitiute the value and find the lapalce inversion.

here,

$\begin{array}{l}{L}^{-1}\left\{\frac{1}{s^3}\right\}=\frac{t^2}{2}\\ \Rightarrow L^{-1}\left\{\frac{1}{{\left(s-5\right)}^3}\right\}=e^{5t}\times \frac{t^2}{2}\\ Therefore,theLaplaceinverseof{L}^{-1}\left\{\frac{1}{{\left(s-5\right)}^3}\right\}\\ \therefore L^{-1}\left\{\frac{1}{{\left(s-5\right)}^3}\right\}=\frac{e^{5t}{t}^2}{2}\end{array}$