<s>[INST] Solve the following differential equations 2x2 with complex eigenvert 
Solve the following equation:
  $$X'=\pmatrix { 2 & 5
  \\ -5 & 8}X$$
  $$x(0)=(5,5)^t$$

I already got out the followings:
$$\lambda_{1,2} = 5 \pm  4i$$
$$v_{1} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} - i\cdot \begin{pmatrix} 4 \\ 0 \end{pmatrix}$$
$$v_{1} = \begin{pmatrix} 3 \\ 5 \end{pmatrix} + i\cdot \begin{pmatrix} 4 \\ 0 \end{pmatrix}$$
And so far I have this(Sry its messy):
$$\begin{align*}\vec{x}_{1}(t) = (3-4i)e^{5t} \begin{pmatrix}a\cos(4t)+b\sin(4t)\end{pmatrix}\end{align*}$$
$$\begin{align*}\vec{x}_{2}(t) = 5\cdot e^{5t} \begin{pmatrix}c\cos(4t)+d\sin(4t)\end{pmatrix}\end{align*}$$
I dont know if this is right so far... but I also tried to sum up x_1:
My attend to solve x_1
This is the right answer in the end:$$\begin{align*}\vec{x}(t) = e^{5t} \begin{pmatrix}5\cos(4t)+\frac{5}{2}\sin(4t)\\ 5\cos(4t)-\frac{5}{2}\sin(4t)\end{pmatrix}\end{align*}$$
tyx
 [/INST] Since we have complex conjugate eigenvalues/eigenvectors, we only need to use one of them.
For $\lambda_1 = 5 + 4 i, v_1 = \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} $, we have
$\begin{align} e^{(5 + 4i)t} \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} \\ = e^{5t}e^{4it} \begin{pmatrix} 3 -4i\\ 5 \end{pmatrix}\end{align} \\
= e^{5t}(\cos 4t + i \sin 4t)\begin{pmatrix} 3 -4i\\ 5 \end{pmatrix} \\
= e^{5t}\begin{pmatrix} 3 \cos 4t + 4 \sin 4t  +i(3 \sin 4t - 4 \cos 4t)\\ 5 \cos 4 t + i( 5  \sin 4t )\end{pmatrix}$
We know that the real and imaginary parts are both solutions, so our general solution is
$$\vec{x}(t) = e^{5t}\left(c_1\begin{pmatrix} 3 \cos 4t + 4 \sin 4t  \\ 5 \cos 4 t \end{pmatrix} + c_2 \begin{pmatrix} 3 \sin 4t - 4 \cos 4t\\  5  \sin 4t \end{pmatrix} \right)$$
We now use our initial condition $x(0)=(5,5)^t$
$$c_1\begin{pmatrix} 3  \\ 5  \end{pmatrix} + c_2 \begin{pmatrix} - 4 \\  0 \end{pmatrix} = \begin{pmatrix} 5  \\ 5  \end{pmatrix}$$
You can use many methods to solve this $2 \times 2$ and it yields $c_1 = 1, c_2 = -\dfrac{1}{2}$ and a final result
$$\begin{align*}\vec{x}(t) = e^{5t} \begin{pmatrix}5\cos 4t+\dfrac{5}{2}\sin 4t\\ 5\cos 4 t-\dfrac{5}{2}\sin 4t\end{pmatrix}\end{align*}$$
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