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

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

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