Question

Determine the inverse Laplace transform of the given function.

$\frac{{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{s}}\mathbf{(}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{2}\mathbf{)}}{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{2}\mathbf{)}}$

Short answer

The solution is$L^{-1}\left\{\frac{e^{-2s}(4s+2)}{(s-1)(s+2)}\right\}=(2e^{t-2}+2e^{-2t+4})u(t-2)$

Step-by-step solution

Step 1:Given Information.

The given function value in s domain is .$\frac{e^{-2s}(4s+2)}{(s-1)(s+2)}$

Determining the inverse Laplace transform

Make partial fraction of the given function, as:

$\frac{4s+2}{(s-1)(s+2)}=\frac{A}{s-1}+\frac{B}{s+2}$

On comparing the two sides of the above equation we get:

$\text{4s+2=A(s+2)+B(s-1)}$

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

$\begin{array}{l}A=2\\ B=2\end{array}$

Thus, the partial fractions are obtained as:

$\frac{4s+2}{(s-1)(s+2)}=\frac{2}{s-1}+\frac{2}{s+2}$

We now have

$\begin{array}{c}{\mathcal{L}}^{-1}\left\{\frac{e^{-2s}(4s+2)}{(s-1)(s+2)}\right\}={\mathcal{L}}^{-1}\left\{\frac{2e^{-2s}}{s-1}\right\}+{\mathcal{L}}^{-1}\left\{\frac{2e^{-2s}}{s+2}\right\}\\ =2e^{t-2}u(t-2)+2e^{-2(t-2)}u(t-2)\\ =(2e^{t-2}+2e^{-2t+4})u(t-2)\end{array}$

Hence, the required inverse Laplace transform is.

$L^{-1}\left\{\frac{{\displaystyle e^{-2s}(4s+2)}}{{\displaystyle (s-1)(s+2)}}\right\}=\left(2e^{t-2}+2e^{-2t+4}\right)u(t-2)$