Use laplace transform to solve the given system a) frac{dx}{dt} -2x- frac{dx}{dt}-y =6e^{3t} 2left(frac{dx}{dt}right)-3x+frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0

Use laplace transform to solve the given system a) frac{dx}{dt} -2x- frac{dx}{dt}-y =6e^{3t} 2left(frac{dx}{dt}right)-3x+frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0

Question
Laplace transform
asked 2021-01-06
Use laplace transform to solve the given system
\(a) \frac{dx}{dt} -2x- \frac{dx}{dt}-y =6e^{3t}\)
\(2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0\)

Answers (1)

2021-01-07
Step 1
Given
\(a) \frac{dx}{dt} -2x- \frac{dx}{dt}-y =6e^{3t}\)
\(2\left(\frac{dx}{dt}\right)-3x+\frac{dx}{dt}-3y=6e^{3t} x(0)=3 y(0)=0\)
We use the Laplace transforms initially,
\(L(x(t))=\bar x(s)\)
\(L(y(t))=\bar y(s)\)
\(L(x'(t))=s\bar x(s)-x(0)\)
\(L(y'(t))=s\bar y(s)-y(0)\)
And in the latter stage, we use the known inverse Laplace transforms to get the required solution.
Step 2
So, we have
\(s\bar x(s)-x(0)-2\bar x(s)-s\bar y(s)+y(0)+\bar y(s)=\frac{6}{(s-3)}\)
\(2s\bar x(s)-2x(0)-3\bar x(s)+s\bar x(s)-y(0)-3\bar y(s)=\frac{6}{(s-3)}\)
Simplifying and applying the initial conditions,
\((s-2)\bar x(s)+(1-s)\bar y(s)=\frac{6}{(s-3)}+3 \dots (1)\)
\((2s-3)\bar x(s)+(x-3)\bar y(s)=\frac{6}{(s-3)}+3 \dots (2)\)
Using Cramer's rule,
\(\bar x(s)=\frac{\begin{vmatrix}\frac{6}{(s-3)}+3 & 1-s \\\frac{6}{(s-3)}+6 & s-3 \end{vmatrix}}{\begin{vmatrix}s-2 & 1-s \\2s-3 & s-3 \end{vmatrix}}\)
\(=\frac{3s-3-\frac{6(s-2)(1-s)}{s-3}}{(s-2)(s-3)-(2s-3)(1-s)}\)
\(=\frac{(s-1)(9s-21)}{3s^2-10s+9} \dots (3)\)
\(=3-\frac{6}{3s^2-10s+9}\)
\(=3-\frac{2}{\left(s-\frac{10}{6}\right)^2+\left(\sqrt{\frac{8}{6}}\right)^2}\)
\(x(t)=3\delta(t)-6/\sqrt{8}e^{\frac{10t}{6}}\sin\left(\frac{\sqrt{8t}}{6}\right)\)
Where, \(\delta(t)\) is the delta function
Step 3
Substituting (3) in (2), we get
\(\frac{(2s-3)(s-1)(9s-21)}{3s^2-10s+9}+(s-3)\bar y(s)=\frac{6}{(s-3)}+6\)
\((s-3)\bar y(s)=\frac{6(s-2)}{(s-3)}-\frac{(2s-3)(s-1)(9s-21)}{3s^2-10s+9}\)
\(\bar y(s)=\frac{6(s-2)}{(s-3)}-\frac{(2s-3)(s-1)(9s-21)}{(s-3)}(3s^2-10s+9)\)
\(\bar y(s)=\frac{-18s^4+159s^3-489s^2+633s-297}{(s-3)^2(3s^2-10s+9)}\)
\(\bar y(s)=-6+\frac{2}{(s-3)^2}+\frac{5-3s}{3s^2-10s+9}\)
\(y(t)=-6\delta(t)+2e^{3t}t-e^{\frac{10t}{6}}\cos\left(\frac{\sqrt8t}{6}\right)\)
Step 4
Conclusion:
\(x(t)=3\delta(t)-\frac{6}{\sqrt8}e^{\frac{10t}{6}}\sin\left(\frac{\sqrt8t}{6}\right)\)
\(y(t)=-6\delta(t)+2e^{3t}t-e^{\frac{10t}{6}}\cos\left(\frac{\sqrt8t}{6}\right)\)
Where, \(\delta(t)\) is the delta function, is the required solution to the system of equations.
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