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

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The discussion focuses on solving a system of differential equations using Laplace Transforms. The equations involve functions x(t) and y(t) with initial conditions set to zero. By applying the Laplace Transform, the transformed equations are manipulated to express Y(s) in terms of s. The final result for the Laplace Transform of y(t) is derived as Y(s) = 54/(s(54 - (s + 1)^2)). The solution appears to be confirmed as correct within the context of the discussion.
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$\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 ) )
 
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