Emeli Hagan

2020-11-23

To find: The solution of the given initial value problem.

Ezra Herbert

Given:
The system of equation is,
$\left(D-4\right)x+6y=9{e}^{-3t}x-\left(D-1\right)y=5{e}^{-3t}$
The initial conditions are given as, $x\left(0\right)=-9y\left(0\right)=4$
Calculation: $LetX\left(s\right)=\mathcal{\text{L}}\left\{x\right\}\left(s\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}Y\left(s\right)=\mathcal{\text{L}}\left\{y\right\}\left(s\right)$
The given differential equation is written as,
${x}^{\prime }-4x+6y=9{e}^{-3t}\dots \left(1\right)$
$x-{y}^{\prime }+y=5{e}^{-3t}\dots \left(2\right)$
Take Laplace transform on both sides of equation (1) and (2) and apply linear property as,
$\mathcal{\text{L}}\left\{{x}^{\prime }-4x+6y\right\}\left(s\right)=\mathcal{\text{L}}\left\{9{e}^{-3t}\right\}\left(s\right)$
$\mathcal{\text{L}}\left\{{x}^{\prime }\right\}\left(s\right)-4\mathcal{\text{L}}\left\{x\right\}\left(s\right)+6\mathcal{\text{L}}\left\{y\right\}\left(s\right)=9\mathcal{\text{L}}\left\{{e}^{-3t}\right\}\left(s\right)\dots \left(3\right)$
$\mathcal{\text{L}}\left\{x-{y}^{\prime }+y\right\}\left(s\right)=\mathcal{\text{L}}\left\{5{e}^{-3t}\right\}\left(s\right)$
$\mathcal{\text{L}}\left\{x\right\}\left(s\right)-\mathcal{\text{L}}\left\{{y}^{\prime }\right\}\left(s\right)+\mathcal{\text{L}}\left\{y\right\}\left(s\right)=5\mathcal{\text{L}}\left\{{e}^{-3t}\right\}\left(s\right)\dots \left(4\right)$
Use Laplace transforms formulas in equation (3) and (4) as,
$sX\left(s\right)-x\left(0\right)-4X\left(s\right)+6Y\left(s\right)=\frac{9}{s+3}$
$X\left(s\right)-sY\left(s\right)-y\left(0\right)+Y\left(s\right)=\frac{5}{s+3}$
Substitute the initial conditions as,
$sX\left(s\right)+9-4X\left(s\right)+6Y\left(s\right)=\frac{9}{s+3}$
$X\left(s\right)-sY\left(s\right)-4+Y\left(s\right)=\frac{5}{s+3}\dots \left(5\right)$
Simplify the system of equations (5) as,
$\left(s-4\right)X\left(s\right)+6Y\left(s\right)=-9+\frac{9}{s+3}\dots \left(6\right)$

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