Use the method of Laplace transformation to solve initial value problem. \frac{dx}{dt}=x-2y , x(0)=-1, y(0)=2 , \frac{dy}{dt}=5x-y

Use the method of Laplace transformation to solve initial value problem.
$\frac{dx}{dt}=x-2y,x\left(0\right)=-1,y\left(0\right)=2$
$\frac{dy}{dt}=5x-y$
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We know in Laplace transformation
$L\left\{{y}^{n}\left(t\right)\right\}={s}^{n}\overline{y}\left(s\right)-{s}^{n-1}y\left(0\right)-{s}^{n-2}{y}^{\prime }\left(0\right)\dots {y}^{n-1}\left(0\right)$
$L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},L\left\{\mathrm{cos}at\right\}=\frac{s}{{s}^{2}+{a}^{2}},L\left\{\mathrm{sin}\left(at\right)\right\}=\frac{a}{{s}^{2}+{a}^{2}}$
Now Given IVP is
$\frac{dx}{dt}=x-2y\left(1\right)x\left(0\right)=-1,y\left(0\right)=2$
$\frac{dy}{dt}=5x-y\left(2\right)$
$⇒{x}^{\prime }\left(t\right)-x+2y=0$
${y}^{\prime }\left(t\right)-5x+y=0$
Taking laplace transformation on both the equation and both the side
From (1)
$L\left\{x\left(t\right)\right\}-L\left\{x\left(t\right)\right\}+2L\left\{y\left(t\right)\right\}=0$
$⇒sx\left(s\right)-x\left(0\right)-x\left(s\right)+2y\left(s\right)=0$
$⇒\left(s-1\right)x\left(s\right)+2y\left(s\right)=-1\left(3\right)$
From (2)
$L\left\{{y}^{\prime }\left(t\right)\right\}-5L\left\{x\left(t\right)\right\}+L\left\{y\left(t\right)\right\}=0$
$⇒sy\left(s\right)-y\left(0\right)-5x\left(s\right)+y\left(s\right)=0$
$⇒\left(s+1\right)y\left(s\right)-5x\left(s\right)=2\left(4\right)$
$⇒5x\left(s-1\right)x\left(s\right)+2y\left(s\right)=-1$
$\left(s-1\right)×\left(s+1\right)y\left(s\right)-5x\left(s\right)=2$
$\left(\left({s}^{2}-1\right)+10\right)y\left(s\right)=2\left(s-1\right)-5$
$⇒y\left(s\right)=\frac{2\left(s-1\right)-5}{{s}^{2}+9}=2\frac{s}{{s}^{2}+9}-\frac{7}{3}\frac{3}{{s}^{2}+9}$
Taking inverse laplace transformation both the side . We get