Question

Use (8) to evaluate inverse Laplace transform.

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

Short answer

The required inverse Laplace transform of the given function is

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

Step-by-step solution

Define the rule (8)

According to (8) we use the following inverse Laplace transform integral:

${\int }_0^{\mathrm{t}}\mathrm{f}\left(\mathrm{T}\right)\mathrm{dT}={\mathrm{L}}^{-1}\left\{\frac{\mathrm{F}\left(\mathrm{s}\right)}{\mathrm{s}}\right\}$

Apply rule (8)

Let $\mathrm{F}\left(\mathrm{s}\right)=\frac{1}{\mathrm{s}\left(\mathrm{s}-1\right)}$

Take inverse Laplace transform on both sides as:

$\begin{array}{l}{L}^{-1}\left\{F\left(s\right)\right\}=L^{-1}\left\{\frac{1}{s\left(s-1\right)}\right\}\\ L^{-1}\left\{\frac{F\left(s\right)}{s}\right\}=L^{-1}\left\{\frac{1}{s^2\left(s-1\right)}\right\}\\ L^{-1}\left\{\frac{F\left(s\right)}{s}\right\}=L^{-1}\left\{\frac{1}{s^{1+1}}\right\}.L^{-1}\left\{\frac{1}{s-1}\right\}\end{array}$

Simplify further as

$\begin{array}{l}{L}^{-1}\left\{\frac{F\left(s\right)}{s}\right\}=t.e^t\\ L^{-1}\left\{\frac{F\left(s\right)}{s}\right\}=f\left(t\right).g\left(t\right)\end{array}$

By convolution theorem, we conclude that$f\left(t\right)=t,g\left(t\right)=e^t$

Using result (8) and convolution theorem, we obtain

$L^{-1}\left\{\frac{1}{s^2\left(s-1\right)}\right\}={\int }_0^{t}t{e}^{t-T}dT$

Apply integration by parts

Simplify${\int }_0^{\mathrm{t}}{\mathrm{te}}^{\mathrm{t}-\mathrm{T}}\mathrm{dT}$using Integrating by parts, as follows:

$\begin{array}{c}{L}^{-1}\left\{\frac{1}{s^2\left(s-1\right)}\right\}={\int }_0^{t}T{e}^{t-T}dT\\ ={\left\{-T{e}^{t-T}\right\}}_0^t+{\int }_0^t{e}^{t-T}dT\\ ={\left\{-T{e}^{t-T}\right\}}_0^t-{\left\{e^{t-T}\right\}}_0^t\end{array}$

Simplify further as:

$\begin{array}{c}{L}^{-1}\left\{\frac{1}{s^2\left(s-1\right)}\right\}=\left\{-t{e}^{t-t}-0\right\}-\left\{e^{t-t}-e^{t-0}\right\}\\ =-t{e}^0-e^0+e^t\\ =e^t-t-1\end{array}$

Thus, the required inverse Laplace transform of the given function is

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