Question

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

16. ${\mathit{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{\left(s^2-5\right)}\right\}\mathbf{=}$……

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

Consider the function,

$L^{-1}\left\{\frac{1}{s^2-5}\right\}$

$\frac{1}{s^2-5}=-\frac{1}{2\sqrt{5}\left(s+\sqrt{5}\right)}+\frac{1}{2\sqrt{5}\left(s-\sqrt{5}\right)}$

Thus,

$L^{-1}\left\{\frac{1}{s^2-5}\right\}=L^{-1}\left\{-\frac{1}{2\sqrt{5}\left(s+\sqrt{5}\right)}+\frac{1}{2\sqrt{5}\left(s-\sqrt{5}\right)}\right\}$

Compute the Laplace inverse

Consider the function,

$L^{-1}\left\{\frac{1}{s^2-5}\right\}=L^{-1}\left\{-\frac{1}{2\sqrt{5}\left(s+\sqrt{5}\right)}+\frac{1}{2\sqrt{5}\left(s-\sqrt{5}\right)}\right\}$

$L^{-1}\left\{a·f\left(s\right)+b·g\left(s\right)\right\}=a·L^{-1}\left\{f\left(s\right)\right\}+b·L^{-1}\left\{g\left(s\right)\right\}$

$L^{-1}\left\{-\frac{1}{2\sqrt{5}\left(s+\sqrt{5}\right)}+\frac{1}{2\sqrt{5}\left(s-\sqrt{5}\right)}\right\}=-L^{-1}\left\{\frac{1}{2\sqrt{5}\left(s+\sqrt{5}\right)}\right\}+L^{-1}\left\{\frac{1}{2\sqrt{5}\left(s-\sqrt{5}\right)}\right\}$

Compute the Laplace inverse of the first term

Consider the function,

$L^{-1}\left\{\frac{1}{2\sqrt{5}\left(s+\sqrt{5}\right)}\right\}$

$\begin{array}{l}=L^{-1}\left\{\frac{1}{2\sqrt{5}}·\frac{1}{s+\sqrt{5}}\right\}\\ =\frac{1}{2\sqrt{5}}{L}^{-1}\left\{\frac{1}{s+\sqrt{5}}\right\}\end{array}$

Here,

$L^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}$

Thus,$L^{-1}\left\{\frac{1}{s+\sqrt{5}}\right\}=e^{-\sqrt{5}t}$

Hence,$L^{-1}\left\{\frac{1}{2\sqrt{5}\left(s+\sqrt{5}\right)}\right\}=\frac{1}{2\sqrt{5}}{e}^{-\sqrt{5}t}$

Compute the Laplace inverse of the second term

Consider,

$L^{-1}\left\{\frac{1}{2\sqrt{5}\left(s-\sqrt{5}\right)}\right\}$

$\begin{array}{l}=L^{-1}\left\{\frac{1}{2\sqrt{5}}·\frac{1}{s-\sqrt{5}}\right\}\\ =\frac{1}{2\sqrt{5}}{L}^{-1}\left\{\frac{1}{s-\sqrt{5}}\right\}\end{array}$

role="math" localid="1663858047603" $\begin{array}{l}{L}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}\\ Thus,\\ L^{-1}\left\{\frac{1}{s-\sqrt{5}}\right\}=e^{\sqrt{5}t}\\ Here,\\ L^{-1}\left\{\frac{1}{2\sqrt{5}\left(s-\sqrt{5}\right)}\right\}=\frac{1}{2\sqrt{5}}{e}^{\sqrt{5}t}\end{array}$

Compute the Laplace inverse

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