# Mahesh's question via email about Laplace Transforms (1)

#### Prove It

##### Well-known member
MHB Math Helper
$\displaystyle x\left( t \right)$ and $\displaystyle y\left( t \right)$ satisfy the following system of differential equations:

$\displaystyle \begin{cases} \frac{\mathrm{d}x}{\mathrm{d}t} + x + 6\,y = 6 \\ \frac{\mathrm{d}y}{\mathrm{d}t} + 9\,x + y = 0 \end{cases}, \quad x \left( 0 \right) = y \left( 0 \right) = 0$

Find the Laplace Transform of $\displaystyle y\left( t \right)$.
Start by taking the Laplace Transform of both equations, which gives

$\displaystyle \begin{cases} s\,X\left( s \right) - s\,x\left( 0 \right) + X\left( s \right) + 6\,Y\left( s \right) = \frac{6}{s} \\ s\,Y\left( s \right) - s\,y\left( 0 \right) + 9\,X\left( s \right) + Y\left( s \right) = 0 \end{cases}$

$\displaystyle \begin{cases} \left( s + 1 \right) X\left( s \right) + 6\,Y\left( s \right) = \frac{6}{s} \\ 9\,X\left( s \right) + \left( s + 1 \right) Y\left( s \right) = 0 \end{cases}$

From the second equation in the system, we have

\displaystyle \begin{align*} 9\,X\left( s \right) &= -\left( s + 1 \right) Y\left( s \right) \\ X\left( s \right) &= -\left( \frac{s + 1}{9} \right) Y\left( s \right) \end{align*}

Substituting into the first equation in the system gives

\displaystyle \begin{align*} \left( s + 1 \right) \left[ -\left( \frac{s + 1}{9} \right) \right] Y\left( s \right) + 6\,Y\left( s \right) &= \frac{6}{s} \\ \left[ 6 -\frac{\left( s + 1 \right) ^2 }{9} \right] Y\left( s \right) &= \frac{6}{s} \\ \left[ \frac{54 - \left( s + 1 \right) ^2 }{9} \right] Y\left( s \right) &= \frac{6}{s} \\ Y\left( s \right) &= \frac{54}{s\left[ 54 - \left( s + 1 \right) ^2 \right]} \end{align*}

In Weblearn this would be entered as

54/( s*( 54 - (s + 1)^2 ) )