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

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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 $y(0)={\begin{bmatrix}-1\\1\end{bmatrix}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

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

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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