Question

# Find x(t) and Y(t) using Laplace transform. begin{cases} frac{dx}{dt}=2x-3y frac{dy}{dt}=y-2x end{cases} x(0)=8 , y(0)=3

Laplace transform
Find x(t) and Y(t) using Laplace transform. $$\begin{cases} \frac{dx}{dt}=2x-3y \frac{dy}{dt}=y-2x \end{cases}$$

$$x(0)=8 , y(0)=3$$

2020-11-10
Step 1 The given pair of equations $$\begin{cases} \frac{dx}{dt}=2x-3y \frac{dy}{dt}=y-2x \end{cases}$$

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