Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 05 (b)

MATH152 April 2010

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Question B 05 (b)

Consider the differential equation system $\mathbf {y} '=A\mathbf {y} ~$ where A has an eigenvalue $\lambda _{1}=1+2i$ with corresponding eigenvector $\mathbf {k} _{1}=\left[{\begin{array}{c}2+3i\\1\end{array}}\right]$ Recall that for real matrices, eigenvalues and eigenvectors come in complex conjugate pairs. Find the real form (no imaginary terms) of the solution that satisfies initial conditions $\mathbf {y} (0)=\left[{\begin{array}{c}1\\0\end{array}}\right]$

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

The two particular solutions from part (a) (each was formed using a pair of eigenvalue and eigenvector of A), i.e. $\mathbf {y} _{1},\mathbf {y} _{2}$ come in complex conjugate pairs too. That means they satisfy $\mathbf {y} _{2}={\bar {\mathbf {y} _{1}}}$. Now, the general complex-valued solution from part (a) is a linear combination of these with arbitrary complex constants $C_{1},C_{2}~$. What form of such constants, then, $C_{1},C_{2}~$ must we choose so that when placed in front of $\mathbf {y} _{1}$ and $\mathbf {y} _{2}$, we can obtain a real-valued solution? You might need to concentrate on the follow fact for complex numbers: $z+{\bar {z}}=2\mathrm {Re} [z]$ is always a real number for any complex number z. Think about how $C_{1},C_{2}~$ should be related to each other.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Continuing from the hint, to choose real-valued solutions, we choose $C_{2}={\bar {C_{1}}}$, so that the general solution becomes: $\mathbf {y} =C_{1}\mathbf {y_{1}} +C_{2}\mathbf {y_{2}} =C_{1}\mathbf {y_{1}} +{\overline {C_{1}\mathbf {y_{1}} }}=2\mathrm {Re} \left[C_{1}\mathbf {y_{1}} \right]$ Recall that $\mathbf {y_{1}} =\mathbf {k_{1}} \exp {\lambda _{1}t}=\left[{\begin{array}{c}2+3i\\1\end{array}}\right]\exp {(1+2i)t}$ Since $\mathrm {Re} \left[\mathbf {y_{1}} \right]=\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}-3\\0\end{array}}\right]e^{t}\sin {2t}$ $\mathrm {Im} \left[\mathbf {y_{1}} \right]=\left[{\begin{array}{c}3\\0\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\sin {2t}$ The general form of real-valued solutions is: ${\begin{aligned}\mathbf {y} &=D_{1}\mathrm {Re} \left[\mathbf {y_{1}} \right]+D_{2}\mathrm {Im} \left[\mathbf {y_{1}} \right]\\&=D_{1}\left(\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}-3\\0\end{array}}\right]e^{t}\sin {2t}\right)\\&+D_{2}\left(\left[{\begin{array}{c}3\\0\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\sin {2t}\right)\end{aligned}}$ where $D_{1}$ and $D_{2}$ are real constants. Their relationship with the previous constants are $D_{1}=2\mathrm {Re} \left[C_{1}\right]$ and $D_{2}=-2\mathrm {Im} \left[C_{1}\right]=e^{t}$ Now if we impose the initial condition $\mathbf {y} (0)=\left[{\begin{array}{c}1\\0\end{array}}\right]$, we get $D_{1}\left[{\begin{array}{c}2\\1\end{array}}\right]+D_{2}\left[{\begin{array}{c}3\\0\end{array}}\right]=\left[{\begin{array}{c}1\\0\end{array}}\right]$ Thus, $D_{1}=0~$ and $D_{2}={\frac {1}{3}}$, and the solution of the initial value problem is ${\frac {1}{3}}\left(\left[{\begin{array}{c}3\\0\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\sin {2t}\right)={\frac {e^{t}}{3}}\left[{\begin{array}{c}3\cos {2t}+2\sin {2t}\\\sin {2t}\end{array}}\right]$ Alternatively, one can solve directly from $\mathbf {y} =2\mathrm {Re} \left[C_{1}\mathbf {y_{1}} \right]$ And impose the initial condition $\mathbf {y} _{0}=\left[{\begin{array}{c}1\\0\end{array}}\right]$ direcly to find that $C_{1}=-{\frac {i}{6}}$. Then compute the real part to get the same result as above. The amount of algebra involved will be slightly less.