Question

Use (8) to evaluate inverse Laplace transform.

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

Short answer

The required inverse Laplace transform of the given function is

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

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 $F\left(s\right)=\frac{1}{s{\left(s-a\right)}^2}$

Now we take inverse Laplace transform on both sides, we get

$\begin{array}{c}{L}^{-1}\left\{F\left(s\right)\right\}=L^{-1}\left\{\frac{1}{{\left(s-a\right)}^2}\right\}\\ L^{-1}\left\{F\left(s\right)\right\}=L^{-1}\left\{{\left.\frac{1}{s^2}\right|}_{s\to s-a}\right\}\\ f\left(t\right)=e^{at}.t\end{array}$

Using the result (8) and convolution theorem, we get

$\begin{array}{l}{L}^{-1}\left\{\frac{1}{s{\left(s-a\right)}^2}\right\}={\int }_0^{t}f\left(T\right)dT\\ L^{-1}\left\{\frac{1}{s{\left(s-a\right)}^2}\right\}={\int }_0^t{e}^{aT}.TdT\end{array}$

Apply integration by parts

Simplify ${\int }_0^t{e}^{aT}.TdT$using Integrating by parts, as follows:

$\begin{array}{c}{L}^{-1}\left\{\frac{1}{s{\left(s-a\right)}^2}\right\}={\int }_0^t{e}^{aT}.TdT\\ ={\left[\frac{e^{aT}.T}{a}\right]}_0^t-\frac{1}{a}{\int }_0^t{e}^{aT}dT\\ =\frac{1}{a}{e}^{at}t-0-\frac{1}{a}{\int }_0^t{e}^{aT}dT\\ =\frac{1}{a}{e}^{at}t-\frac{1}{a^2}{\left[e^{aT}\right]}_0^t\end{array}$

Simplify further as follows:

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

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

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