# Science:Math Exam Resources/Courses/MATH152/April 2022/Question B4 (b)/Solution 1

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 ${\textstyle e^{\lambda _{1}t}{\vec {v}}_{1}}$:

${\displaystyle e^{\lambda _{1}t}{\vec {v}}_{1}=e^{(2+i)t}\left[{\begin{array}{c}1\\-1+i\end{array}}\right]=e^{2t}(\cos(t)+i\sin(t))\left[{\begin{array}{c}1\\-1+i\end{array}}\right]=e^{2t}\left(\left[{\begin{array}{c}\cos(t)\\-\cos(t)-\sin(t)\end{array}}\right]+i\left[{\begin{array}{c}\sin(t)\\\cos(t)-\sin(t)\end{array}}\right]\right).}$

The complex conjugate yields the other part of the solution:

${\displaystyle e^{\lambda _{2}t}{\vec {v}}_{2}=e^{{\overline {\lambda _{1}}}t}{\overline {{\vec {v}}_{1}}}={\overline {e^{\lambda _{1}t}{\vec {v}}_{1}}}=e^{2t}\left(\left[{\begin{array}{c}\cos(t)\\-\cos(t)-\sin(t)\end{array}}\right]-i\left[{\begin{array}{c}\sin(t)\\\cos(t)-\sin(t)\end{array}}\right]\right).}$

Then, to get the general form, we take the real and imaginary parts of the general complex solution ${\displaystyle c_{1}e^{\lambda _{1}t}{\vec {v}}_{1}+c_{2}e^{\lambda _{2}t}{\vec {v}}_{2}}$:

${\textstyle {\vec {y}}(t)=(c_{1}+c_{2})e^{2t}\left[{\begin{array}{c}\cos(t)\\-\cos(t)-\sin(t)\end{array}}\right]+(c_{1}-c_{2})e^{2t}\left[{\begin{array}{c}\sin(t)\\\cos(t)-\sin(t)\end{array}}\right],}$

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 ${\displaystyle c_{1},c_{2}}$ are arbitrary.