91Ó°ÊÓ

When specific initial conditions are supplied, especially when the initial values are zero, the Laplace transform is a handy method of solving certain types of differential equations. Laplace transform$\mathcal{L}$ of a function$f\left(t\right)$ is defined as:

$\mathcal{L}\left\{f\right(t\left)\right\}={∫}_0^{\frac{<}{>}}{e}^{-st}f\left(t\right)dt$

In words, we can describe this expression as the Laplace transform of f(t) equals function F of s, that is, $\mathcal{L}\left\{f\right(t\left)\right\}=F\left(s\right)$.

Given that $y\text{'}\text{'}\left(t\right)+5y\text{'}\left(t\right)+6y\left(t\right)=g\left(t\right),t>0$.

Take Laplace transform on both sides as:

$\begin{array}{rcl}\mathcal{L}\left[y\text{'}\text{'}\left(t\right)+5y\text{'}\left(t\right)+6y\left(t\right)\right]\left(s\right)& =& \mathcal{L}\left[g\right(t\left)\right]\left(s\right)\\ s^{2}Y\left(s\right)-sY\left(0\right)-Y\text{'}\left(0\right)+5\left[sY\right(s)-Y(0\left)\right]+6Y\left(s\right)& =& G\left(s\right)\end{array}$

Simplify the equation using all initial conditions are zero as follows:

$\begin{array}{rcl}{s}^{2}Y\left(s\right)+5sY\left(s\right)+6Y\left(s\right)& =& G\left(s\right)\\ \left(s^2+5s+6\right)Y\left(s\right)& =& G\left(s\right)\\ \frac{Y\left(s\right)}{G\left(s\right)}& =& \frac{1}{s^2+5s+6}\\ \frac{Y\left(s\right)}{G\left(s\right)}& =& \frac{1}{(s+3)(s+2)}\end{array}$

Hence, the value of transfer function is$\frac{1}{(s+3)(s+2)}$.