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
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asked 2020-11-09
Find x(t) and Y(t) using Laplace transform. \(\begin{cases} \frac{dx}{dt}=2x-3y
\frac{dy}{dt}=y-2x \end{cases}\)

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

Answers (1)

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, \(x(t)=5e^{-t}+3e^{4t}\)
\(y(t)=5e^{-t}-2e^{4t}\)
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