Question

Determine the inverse Laplace transform of the given function.

$\frac{\mathbf{2}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{s}\mathbf{-}\mathbf{1}}{{\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{2}\mathbf{)}}$

Short answer

Therefore, the solution is

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

Step-by-step solution

Given Information

The given function value in s domain is

$\frac{2s^2+3s-1}{{(s+1)}^2(s+2)}$

Use partial fractions

Make partial fraction of the given function, as:

$\begin{array}{c}\frac{2s^2+3s-1}{{(s+1)}^2(s+2)}=\frac{A}{s+1}+\frac{B}{{(s+1)}^2}+\frac{C}{s+2}\\ =\frac{A(s+1)(s+2)+B(s+2)+C{(s+1)}^2}{{(s+1)}^2(s+2)}\end{array}$

On comparing the coefficients with $s=-1,-2,0$respectively we get

$\begin{array}{c}A=1\\ B=-2\\ C=1\end{array}$

Thus, the partial fractions are obtained as:

$\frac{2s^2+3s-1}{{(s+1)}^2(s+2)}=\frac{1}{s+1}-\frac{2}{{(s+1)}^2}+\frac{1}{s+2}$

Take Inverse Laplace transform

Take inverse Laplace transform using ${\mathcal{L}}^{-1}\left\{\frac{{\displaystyle \text{1}}}{{\displaystyle s-a}}\right\}\left(t\right)=e^{at}$and ${\mathcal{L}}^{-1}\left\{\frac{{\displaystyle n!}}{{\displaystyle {(s-a)}^{n+1}}}\right\}\left(t\right)=t^n{e}^{at}$as:

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

Hence, the required inverse Laplace transform is

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