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

MATH152 April 2010

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[hide] 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 $\mathbf {y} (0)=\left[{\begin{array}{c}1\\0\end{array}}\right]$

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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!

[show] 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.

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[show] 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.