Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 05 (b)/Solution 1

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.