Q14RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the inverse Laplace transform of the given function.

$\frac{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{16}\mathbf{s}\mathbf{+}\mathbf{9}}{\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{+}\mathbf{3}\mathbf{)}\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{2}\mathbf{)}}$

Step-by-Step Solution

Answer

Therefore, the solution is

${\mathcal{L}}^{-1}\left\{\frac{s^2+16s+9}{(s+1)(s+3)(s-2)}\right\}\left(t\right)=e^{-t}-3e^{-3t}+3e^{2t}$

Step 1: Given Information

The given function value in s domain is $\frac{s^2+16s+9}{(s+1)(s+3)(s-2)}$

Step 2: Use partial fractions

Make partial fraction of the given function, as:

$\begin{array}{c}\frac{s^2+16s+9}{(s+1)(s+3)(s-2)}=\frac{A}{s+1}+\frac{B}{s+3}+\frac{C}{s-2}\\ =\frac{A(s+3)(s-2)+B(s+1)(s-2)+C(s+1)(s+3)}{(s+1)(s+3)(s-2)}\end{array}$

On comparing the coefficients with $s=-1,-3,2$respectively we get

$\begin{array}{l}A=1\\ B=-3\\ C=3\end{array}$

Thus, the partial fractions are obtained as:

$\frac{s^2+16s+9}{(s+1)(s+3)(s-2)}=\frac{1}{s+1}-\frac{3}{s+3}+\frac{3}{s-2}$

Step 3: Take Inverse Laplace transform

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

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

Hence, the required inverse Laplace transform is

${\mathcal{L}}^{-1}\left\{\frac{s^2+16s+9}{(s+1)(s+3)(s-2)}\right\}\left(t\right)=e^{-t}-3e^{-3t}+3e^{2t}$