\begin{equation} \begin{cases} x_1’ = 5x_1 - 5x_2 \\ x_2’ = 2x_1 -x_2 \end{cases} \end{equation}
This gives rise to:
\begin{equation} A = \mqty(5 & -5 \\ 2 &-1) \end{equation}
Solving the characteristic polynomial gives:
\begin{equation} (5-\lambda)(-1-\lambda) + 10 = \lambda^{2} - 4\lambda +5 \end{equation}
Therefore, our solutions are imaginary!
\begin{equation} \lambda_{1}, \lambda_{2} = 2 \pm i \end{equation}
Aside: we only need to deal with one
Notably, anything that satisfies the original polynomial, its conjugates also satisfies:
\begin{equation} \bar{\lambda^{2}-4\lambda +5} = 0= {\bar{\lambda}}^{2} - 4\bar{\lambda} + 5 \end{equation}
Further, for some:
\begin{equation} Av = \lambda v \end{equation}
we have:
\begin{equation} A \bar{v} = \lambda \bar{v} \end{equation}
meaning if we just figured the eigenvector of one of the lambdas we are good
Now, let us consider the case before with \(\lambda = 2 +i\). We therefore have:
\begin{equation} \mqty(3-i & -5 \\ 2 & -3-i) \mqty(a \\ b) = \mqty(0 \\ 0) \end{equation}
This gives one particular null space, such as:
\begin{equation} v = \mqty(5 \\ 3-i) \end{equation}
This gives rise to:
\begin{equation} u’ = (2+i)u \end{equation}
which means:
\begin{equation} u(t) = ce^{(2+i)t} \end{equation}
finally, resulting in:
\begin{equation} x(t) = ce^{(2+i)t} \mqty(5 \\ 3-i) \end{equation}
which is a particular solution. Now, the general solution would tack on a complex conjugate, which doesn’t actually add any new information.
Instead, we can actually use Euler to break this into two, independent, and equally valid solutions:
\begin{equation} x(t) = e^{2t} \qty(\cos t + i \sin t) \qty( \mqty(5 \\3) - i \mqty(0 \\ 1)) \end{equation}
finally, we obtain:
\begin{equation} x(t) = e^{2t} \qty( \cos t \mqty(5 \\ 1) + \sin t \mqty(0 \\1)) + i e^{2t} \qty( \sin t \mqty(5 \\ 1) - \cos t \mqty(0 \\1)) \end{equation}
each of which individual is a solution:
\begin{equation} x_1(t) =e^{2t} \qty( \cos t \mqty(5 \\ 1) + \sin t \mqty(0 \\1)) \end{equation}
and:
\begin{equation} x_2(t) = e^{2t} \qty( \sin t \mqty(5 \\ 1) - \cos t \mqty(0 \\1)) \end{equation}