 # Use laplace transform to solve the given system a) frac{dx}{dt} -2x- frac{dx}{dt}-y =6e^{3t} 2left(frac{dx}{dt}right)-3x+frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0 amanf 2021-01-06 Answered
Use laplace transform to solve the given system
$a\right)\frac{dx}{dt}-2x-\frac{dx}{dt}-y=6{e}^{3t}$
$2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6{e}^{3t}x\left(0\right)=3y\left(0\right)=0$
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Step 1
Given
$a\right)\frac{dx}{dt}-2x-\frac{dx}{dt}-y=6{e}^{3t}$
$2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6{e}^{3t}x\left(0\right)=3y\left(0\right)=0$
We use the Laplace transforms initially,
$L\left(x\left(t\right)\right)=\overline{x}\left(s\right)$
$L\left(y\left(t\right)\right)=\overline{y}\left(s\right)$
$L\left({x}^{\prime }\left(t\right)\right)=s\overline{x}\left(s\right)-x\left(0\right)$
$L\left({y}^{\prime }\left(t\right)\right)=s\overline{y}\left(s\right)-y\left(0\right)$
And in the latter stage, we use the known inverse Laplace transforms to get the required solution.
Step 2
So, we have
$s\overline{x}\left(s\right)-x\left(0\right)-2\overline{x}\left(s\right)-s\overline{y}\left(s\right)+y\left(0\right)+\overline{y}\left(s\right)=\frac{6}{\left(s-3\right)}$
$2s\overline{x}\left(s\right)-2x\left(0\right)-3\overline{x}\left(s\right)+s\overline{x}\left(s\right)-y\left(0\right)-3\overline{y}\left(s\right)=\frac{6}{\left(s-3\right)}$
Simplifying and applying the initial conditions,
$\left(s-2\right)\overline{x}\left(s\right)+\left(1-s\right)\overline{y}\left(s\right)=\frac{6}{\left(s-3\right)}+3\dots \left(1\right)$
$\left(2s-3\right)\overline{x}\left(s\right)+\left(x-3\right)\overline{y}\left(s\right)=\frac{6}{\left(s-3\right)}+3\dots \left(2\right)$
Using Cramer's rule,
$\overline{x}\left(s\right)=\frac{|\begin{array}{cc}\frac{6}{\left(s-3\right)}+3& 1-s\\ \frac{6}{\left(s-3\right)}+6& s-3\end{array}|}{|\begin{array}{cc}s-2& 1-s\\ 2s-3& s-3\end{array}|}$
$=\frac{3s-3-\frac{6\left(s-2\right)\left(1-s\right)}{s-3}}{\left(s-2\right)\left(s-3\right)-\left(2s-3\right)\left(1-s\right)}$
$=\frac{\left(s-1\right)\left(9s-21\right)}{3{s}^{2}-10s+9}\dots \left(3\right)$
$=3-\frac{6}{3{s}^{2}-10s+9}$
$=3-\frac{2}{{\left(s-\frac{10}{6}\right)}^{2}+{\left(\sqrt{\frac{8}{6}}\right)}^{2}}$
$x\left(t\right)=3\delta \left(t\right)-6/\sqrt{8}{e}^{\frac{10t}{6}}\mathrm{sin}\left(\frac{\sqrt{8t}}{6}\right)$
Where, $\delta \left(t\right)$ is the delta function
Step 3
Substituting (3) in (2), we get
$\frac{\left(2s-3\right)\left(s-1\right)\left(9s-21\right)}{3{s}^{2}-10s+9}+\left(s-3\right)\overline{y}\left(s\right)=\frac{6}{\left(s-3\right)}+6$

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