Science:Math Exam Resources/Courses/MATH152/April 2022/Question B4 (b)

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MATH152 April 2022

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Question B4 (b)
Find the general solution of the differential equation ${\textstyle {\frac {d}{dt}}{\vec {y}}(t)=A{\vec {y}}(t)}$ in the real form.

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
Recall that the general solution to a system of differential equations of this form is ${\textstyle {\vec {y}}(t)=c_{1}e^{\lambda _{1}t}{\vec {v}}_{1}+c_{2}e^{\lambda _{2}t}{\vec {v}}_{2}}$ , where ${\vec {v}}_{1},{\vec {v}}_{2}$ are eigenvectors associated to the eigenvalues $\lambda _{1},\lambda _{2}$ . The real and imaginary parts of these complex solutions are two real solutions.

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Solution
The solution to a ${\textstyle 2\times 2}$ system of differential equations whose eigenvalues form a complex conjugate pair only requires writing out the solution for one of the eigenvalues, since we can quickly obtain the other by complex conjugation. We start with ${\textstyle e^{\lambda _{1}t}{\vec {v}}_{1}}$ : $e^{\lambda _{1}t}{\vec {v}}_{1}=e^{(2+i)t}\left[{\begin{array}{c}1\\-1+i\end{array}}\right]=e^{2t}(\cos(t)+i\sin(t))\left[{\begin{array}{c}1\\-1+i\end{array}}\right]=e^{2t}\left(\left[{\begin{array}{c}\cos(t)\\-\cos(t)-\sin(t)\end{array}}\right]+i\left[{\begin{array}{c}\sin(t)\\\cos(t)-\sin(t)\end{array}}\right]\right).$ The complex conjugate yields the other part of the solution: $e^{\lambda _{2}t}{\vec {v}}_{2}=e^{{\overline {\lambda _{1}}}t}{\overline {{\vec {v}}_{1}}}={\overline {e^{\lambda _{1}t}{\vec {v}}_{1}}}=e^{2t}\left(\left[{\begin{array}{c}\cos(t)\\-\cos(t)-\sin(t)\end{array}}\right]-i\left[{\begin{array}{c}\sin(t)\\\cos(t)-\sin(t)\end{array}}\right]\right).$ Then, to get the general form, we take the real and imaginary parts of the general complex solution $c_{1}e^{\lambda _{1}t}{\vec {v}}_{1}+c_{2}e^{\lambda _{2}t}{\vec {v}}_{2}$ : ${\textstyle {\vec {y}}(t)=(c_{1}+c_{2})e^{2t}\left[{\begin{array}{c}\cos(t)\\-\cos(t)-\sin(t)\end{array}}\right]+(c_{1}-c_{2})e^{2t}\left[{\begin{array}{c}\sin(t)\\\cos(t)-\sin(t)\end{array}}\right],}$ which we may rewrite as ${\textstyle {\vec {y}}(t)=c_{1}e^{2t}\left[{\begin{array}{c}\cos(t)\\-\cos(t)-\sin(t)\end{array}}\right]+c_{2}e^{2t}\left[{\begin{array}{c}\sin(t)\\\cos(t)-\sin(t)\end{array}}\right],}$ since the constants are $c_{1},c_{2}$ are arbitrary.