Science:Math Exam Resources/Courses/MATH152/April 2013/Question B 06 (c)

MATH152 April 2013

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Question B 06 (c)
Consider the system of linear differential equations given by ${\begin{aligned}y'_{1}(t)&=3y_{1}(t)-2y_{2}(t)\\y'_{2}(t)&=4y_{1}(t)-y_{2}(t)\end{aligned}}$ Find the solution of the system of linear differential equations above with the initial condition $y(0)={\begin{bmatrix}-1\\1\end{bmatrix}}$

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Hint
To incorporate the initial value $\displaystyle y(0)={\begin{bmatrix}-1\\1\end{bmatrix}}$, the constants $\displaystyle C_{1},C_{2}$ must be chosen such that (from part (b)): $y(t)=C_{1}e^{t}{\begin{bmatrix}\cos(2t)\\\cos(2t)+\sin(2t)\end{bmatrix}}+C_{2}e^{t}{\begin{bmatrix}\sin(2t)\\\sin(2t)-\cos(2t)\end{bmatrix}}$ holds for $\displaystyle t=0$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Using the solution from part (b), $y(t)=C_{1}e^{t}{\begin{bmatrix}\cos(2t)\\\cos(2t)+\sin(2t)\end{bmatrix}}+C_{2}e^{t}{\begin{bmatrix}\sin(2t)\\\sin(2t)-\cos(2t)\end{bmatrix}}$ we have for $\displaystyle t=0$ $y(0)=C_{1}{\begin{bmatrix}1\\1\end{bmatrix}}+C_{2}{\begin{bmatrix}0\\-1\end{bmatrix}}$ Because the given intial values are $\displaystyle y(0)={\begin{bmatrix}-1\\1\end{bmatrix}}$, we write ${\begin{bmatrix}-1\\1\end{bmatrix}}=C_{1}{\begin{bmatrix}1\\1\end{bmatrix}}+C_{2}{\begin{bmatrix}0\\-1\end{bmatrix}}$ This is ${\begin{aligned}-1&=C_{1}\\1&=C_{1}-C_{2}\end{aligned}}$ and $\displaystyle C_{1}=-1$ and $\displaystyle C_{2}=-2$. Hence, the solution is ${\begin{aligned}y(t)&=-e^{t}{\begin{bmatrix}\cos(2t)\\\cos(2t)+\sin(2t)\end{bmatrix}}-2e^{t}{\begin{bmatrix}\sin(2t)\\\sin(2t)-\cos(2t)\end{bmatrix}}\\&=e^{t}{\begin{bmatrix}-2\sin(2t)-\cos(2t)\\-3\sin(2t)+\cos(2t)\end{bmatrix}}\end{aligned}}$

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