Q13RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 3-10, determine the Laplace transform of the given function.$\frac{\mathbf{4}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{13}\mathbf{s}\mathbf{+}\mathbf{19}}{\mathbf{(}\mathbf{s}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{(}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{s}\mathbf{+}\mathbf{13}\mathbf{)}}$

Step-by-Step Solution

Answer

Therefore, the solution is${\mathcal{L}}^{-1}\left\{\frac{4s^2+13s+19}{(s-1)(s^2+4s+13)}\right\}\left(t\right)=2e^t+2e^{-2t}\cos 3t+e^{-2t}\sin 3t$

Step 1: Given Information

The given value is$\frac{4s^2+13s+19}{(s-1)(s^2+4s+13)}$

Step 2: Use partial fractions

Make partial fraction of the given function, as:

$\begin{array}{c}\frac{4s^2+13s+19}{(s-1)(s^2+4s+13)}=\frac{4s^2+13s+19}{(s-1)({(s+2)}^2+3^2)}\\ =\frac{A}{s-1}+\frac{B(s+2)+3C}{{(s+2)}^2+3^2}\end{array}$

On comparing the numerator of above equation we get:

$4s^2+13s+19=(A+B)s^2+(4A+B+3C)s+13A-2B-3C$

On comparing the coefficients, we get the system

$\{\begin{array}{l}A+B=4\\ 4A+B+3C=13\\ 13A-2B-3C=19\end{array}$

On simplifying further, we get

 $\{\begin{array}{l}A=4-B\\ C=B-1\\ -18B=-36\end{array}$

On solving the equations, we get

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

Thus, the partial fractions are obtained as:

$\frac{4s^2+13s+19}{(s-1)(s^2+4s+13)}=\frac{2}{s-1}+\frac{2(s+2)+3}{{(s+2)}^2+3^2}$

Step 3: Take Inverse Laplace transform

Take inverse Laplace transform using ${\mathcal{L}}^{-1}\left\{\frac{s-a}{{(s-a)}^2+b^2}\right\}\left(t\right)=e^{at}\cos bt$and ${\mathcal{L}}^{-1}\left\{\frac{b}{{(s-a)}^2+b^2}\right\}\left(t\right)=e^{at}\sin bt$as:

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

Hence, the required inverse Laplace transform is

${\mathcal{L}}^{-1}\left\{\frac{4s^2+13s+19}{(s-1)(s^2+4s+13)}\right\}\left(t\right)=2e^t+2e^{-2t}\cos 3t+e^{-2t}\sin 3t$