Question

Use (8) to evaluate inverse Laplace transform.

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

Short answer

The required inverse Laplace transform of the given function is

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

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

Take inverse Laplace transform, using ${\mathrm{L}}^{-1}\left\{\frac{1}{\mathrm{s}-\mathrm{a}}\right\}={\mathrm{e}}^{\mathrm{at}}$on both sides,

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

Using result (8), we get

$\begin{array}{c}{L}^{-1}\left\{\frac{1}{s\left(s-1\right)}\right\}={\int }_0^t{e}^{T}dT\\ L^{-1}\left\{\frac{1}{s\left(s-1\right)}\right\}={\left\{e^T\right\}}_0^t\\ L^{-1}\left\{\frac{1}{s\left(s-1\right)}\right\}=\left\{e^t-e^0\right\}\\ L^{-1}\left\{\frac{1}{s\left(s-1\right)}\right\}=e^t-1\end{array}$

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

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