The solution to a
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
:
![{\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).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/4c330a482fbf0769dc7363acecc847c35c2e7bcb)
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).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d2abb63aadbe8a2225edec31138ed5587a81b8dc)
Then, to get the general form, we take the real and imaginary parts of the general complex solution
:
which we may rewrite as
since the constants are
are arbitrary.