91Ó°ÊÓ

Let .$f\left(z\right)=\frac{z}{\left(z+1\right)\left(z^2+4\right)}$

$∴f\left(z\right)e^{zt}=f\left(z\right)=\frac{z{e}^{zt}}{\left(z+1\right)\left(z^2+4\right)}$

Now,

$\begin{array}{rcl}f\left(z\right)& =& \frac{z}{\left(z+1\right)\left(z^2+4\right)}\\ & =& \frac{z}{\left(z+1\right)\left(z-2i\right)\left(z+2i\right)}\end{array}$

The four poles of the function are,

$z= -1,  2i,  -2i$

TheResult 7.16 is as follows:

f(t) = Sum of residues of ${\mathrm{e}}^{\mathrm{zt}}\mathrm{f}\left(\mathrm{z}\right)$.

The residues at these poles are as follows:

At : z = -1

$\begin{array}{rcl}res\left(f\left(z\right)e^{zt}|z=-1\right)& =& \underset{z→-1}{\lim }\frac{z{e}^{zt}}{\left(z-2i\right)\left(z+2i\right)}\\ & =& \frac{-e^{-t}}{\left(-1-2i\right)\left(-1+2i\right)}\\ & =& -\frac{e^{-t}}{5}\end{array}$

At : z = 2i

$\begin{array}{rcl}res\left(f\left(z\right)e^{zt}|z=2i\right)& =& \underset{z→2i}{\lim }\frac{z{e}^{zt}}{\left(z+1\right)\left(z+2i\right)}\\ & =& \frac{2i{e}^{2it}}{\left(2i+1\right)\left(2i+2i\right)}\\ & =& \frac{2i{e}^{2it}}{\left(2i+1\right)4i}\\ & =& \frac{e^{2it}}{2\left(2i+1\right)}\end{array}$

At : z = -2i

$\begin{array}{rcl}res\left(f\left(z\right)e^{zt}|z=-2i\right)& =& \underset{z→2i}{\lim }\frac{z{e}^{zt}}{\left(z+1\right)\left(z-2i\right)}\\ & =& \frac{-2i{e}^{-2it}}{\left(-2i+1\right)\left(-2i-2i\right)}\\ & =& \frac{-2i{e}^{-2it}}{\left(-2i+1\right)\left(-4i\right)}\\ & =& \frac{e^{-2it}}{2\left(-2i+1\right)}\end{array}$

Hence, the function f(t) is given by,

$\begin{array}{rcl}f\left(t\right)& =& res\left(f\left(z\right)e^{zt}|z=-1\right)+res\left(f\left(z\right)e^{zt}|z=2i\right)+res\left(f\left(z\right)e^{zt}|z=-2i\right)\\ & =& -\frac{e^{-t}}{5}+\frac{e^{2it}}{2\left(2i+1\right)}+\frac{e^{-2it}}{2\left(-2i+1\right)}\\ & =& -\frac{e^{-t}}{5}+\frac{1}{5}\left(\frac{e^{2it}+e^{-2it}}{2}\right)+\frac{2}{5}\left(\frac{e^{2it}-e^{-2it}}{2i}\right)\\ & =& \frac{2\sin \left(2t\right)+\cos \left(2t\right)-e^{-t}}{5}\end{array}$

$∴f\left(t\right)=\frac{2\sin \left(2t\right)+\cos \left(2t\right)-e^{-t}}{5}$

.

Hence, the inverse Laplace transform of the given function is,

$f\left(t\right)=\frac{2\sin \left(2t\right)+\cos \left(2t\right)-e^{-t}}{5}$