Question

Use (8) to evaluate inverse Laplace transform.

${\mathbf{L}}^{\mathbf{-}\mathbf{1}}\left\{\frac{1}{s^3\left(s-1\right)}\right\}$

Short answer

The required inverse Laplace transform of the given function is

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

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

Now by using partial fraction method we decompose the fraction.

$\begin{array}{c}F\left(s\right)=\frac{1}{s^2\left(s-1\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s-1}\\ 1=As\left(s-1\right)+B\left(s-1\right)+C{s}^2\\ 1=A{s}^2-As+B{s}^2-B+C{s}^2\\ 1=\left(A+C\right)s^2-As-B\end{array}$

Comparing the coefficients and solving, we get

$B=-1,A=-1,C=1$

The partial fractions are written as follows:

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

Take inverse Laplace

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

$\begin{array}{c}{L}^{-1}\left\{F\left(s\right)\right\}=L^{-1}\left\{\frac{-1}{s}+\frac{-1}{s^2}+\frac{1}{s-1}\right\}\\ L^{-1}\left\{\frac{1}{s^3\left(s-1\right)}\right\}={\int }_0^t\left(-1-T+e^T\right)dT\\ ={\left[-T\right]}_0^t-{\left[\frac{T^2}{2}\right]}_0^t+{\left[e^T\right]}_0^t\\ =\left[-t+0\right]-\left[\frac{t^2}{2}-0\right]+e^t-e^0\\ =-t-\frac{t^2}{2}+e^t-1\end{array}$

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

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