91Ó°ÊÓ

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

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:

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

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

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