Homogenous Equation with Complex Vector Solution: Converting to Real Functions Solving the system
$$
\begin{array}{l}\frac{d x}{d t}=6 x-y \\ \frac{d y}{d t}=5 x+4 y\end{array}
$$
we get $\lambda_{1}=5+2 i, \lambda_{2}=5-2 i$ eigenvalues. So eigenvectors and corresponding solutions is:
$$
\mathbf{K}_{1}=\left(\begin{array}{c}1 \\ 1-2 i\end{array}\right), \quad \mathbf{X}_{1}=\left(\begin{array}{c}1 \\ 1-2 i\end{array}\right) e^{(5+2 i) t}
$$
$$
\mathbf{K}_{2}=\left(\begin{array}{c}1 \\ 1+2 i\end{array}\right), \quad \mathbf{X}_{2}=\left(\begin{array}{c}1 \\ 1+2 i\end{array}\right) e^{(5-2 i) t}
$$
Thus general solution:
$$
\mathbf{X}=c_{1}\left(\begin{array}{c}1 \\ 1-2 i\end{array}\right) e^{(5+2 i) t}+c_{2}\left(\begin{array}{c}1 \\ 1+2 i\end{array}\right) e^{(5-2 i) t}
$$
So converting the above solution to a real function via Euler's:
$$
\begin{array}{l}e^{(5+2 i) t}=e^{5 t} e^{2 t i}=e^{5 t}(\cos 2 t+i \sin 2 t) \\ e^{(5-2 i) t}=e^{5 t} e^{-2 t i}=e^{5 t}(\cos 2 t-i \sin 2 t)\end{array}
$$
My text states to collect terms and replace $ c_1 + c_2 $ by $ C_1 $ and $ (c_1 - c_2)i $ by $ C_2 $ the solution becoming $\mathbf{X}=C_{1} \mathbf{X}_{1}+C_{2} \mathbf{X}_{2}$ where
$$
\mathbf{X}_{1}=\left[\left(\begin{array}{l}1 \\ 1\end{array}\right) \cos 2 t-\left(\begin{array}{r}0 \\ -2\end{array}\right) \sin 2 t\right] e^{5 t}
$$
$$
\mathbf{X}_{2}=\left[\left(\begin{array}{r}0 \\ -2\end{array}\right) \cos 2 t+\left(\begin{array}{l}1 \\ 1\end{array}\right) \sin 2 t\right] e^{5 t}
$$
I seem to be making a mistake somewhere and don't get to the same simplification. I'm looking for some of the steps to the books specification. I've tried a few times by multiplying the general solution across rows, fully expanding outside of its matrix form.

We have
$$\lambda_1 = 5 + 2 i, v_1 = \begin{pmatrix} 1 \\ 1 - 2 i \end{pmatrix} = \begin{pmatrix} 1 \\ 1  \end{pmatrix} +i \begin{pmatrix} 0 \\ - 2  \end{pmatrix}$$
We form $x_1(t) = e^{\lambda_1 t}v_1$, while expanding using Euler's Formula and this approach
$$x_1(t) = e^{5t}(\cos 2 t + i \sin 2t) \left(\begin{pmatrix} 1 \\ 1  \end{pmatrix} +i \begin{pmatrix} 0 \\ - 2  \end{pmatrix} \right) = e^{5t}\left(\begin{pmatrix} 1 \\ 1  \end{pmatrix}\cos 2t +i^2 \begin{pmatrix} 0 \\ - 2  \end{pmatrix}\sin 2t \right)$$
This is enough to solve the system, but you can try the second eigenvector.
