# Find x(t) and Y(t) using Laplace transform. (dx)/(dt)=2x-3y (dy)/(dt)=y-2x x(0)=8 , y(0)=3

Find x(t) and Y(t) using Laplace transform. $\left\{\begin{array}{l}\frac{dx}{dt}=2x-3y\frac{dy}{dt}=y-2x\end{array}$
$x\left(0\right)=8,y\left(0\right)=3$

You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

liannemdh
Step 1 The given pair of equations $\left\{\begin{array}{l}\frac{dx}{dt}=2x-3y\frac{dy}{dt}=y-2x\end{array}$
$x\left(0\right)=8,y\left(0\right)=3$ Step 2 Take Laplace transform $sx-x\left(0\right)=2x-3y$
$\left(s-2\right)x=8-3y$

$sy-y\left(0\right)=y-2x$
Put equation (1) in (2) Step 3 $\left(s-1\right)×\left(\frac{8\left(s-2\right)x}{3}\right)=3-2x$
$8\left(s-1\right)-\left(s-1\right)\left(s-2\right)x=9-6x$
$\left(\left(s-1\right)\left(s-2\right)-6\right)x=8\left(s-1\right)-9$
$x=\frac{8s-17}{{s}^{2}-3s-4}$
$x=\frac{8s-17}{\left(s-4\right)\left(s+1\right)}$ Take partial fractions $x=\frac{3}{s-4}+\frac{5}{s+1}$ Take inverse Laplace $x\left(t\right)=5{e}^{-t}+3{e}^{4t}$ Step 4 $\frac{dx}{dt}=2x-3y$
$y=\frac{10{e}^{-t}+6{e}^{4t}-\left(-5{e}^{-t}+12{e}^{4t}\right)}{3}=\frac{15{e}^{-t}-6{e}^{4t}}{3}$
$y\left(t\right)=5{e}^{-t}-2{e}^{4t}$
Therefore, $x\left(t\right)=5{e}^{-t}+3{e}^{4t}$
$y\left(t\right)=5{e}^{-t}-2{e}^{4t}$