Solving a systemof eqns by given initial conditions I have the system
$x'(t)=2x+8y$
$y'(t)=-x-2y$
Which has the I.C. $X(0)=(6,-2)$. So I take that it means this:
$6=2x+8y$
$-2=-x-2y$
and then it is solved as any other system? We get eigenvector $(1,1/2)$, but since  the eigenvalues of the system are two, $\pm 2i$, we have to find the second generalized eigenvector.
$$ \left( \begin{array}{cc}
2-2i & 8 \\
-1 & -2-2i
\end{array} \right)
%
\left( \begin{array}{cc}
1\\
\frac{1}{2}
\end{array} \right)=%
\left( \begin{array}{cc}
6-2i \\
-2-i
\end{array} \right)
$$
So the second vector would be $(6-2i, -2-i)$. Using the general solutions  for the system, we get
\begin{equation}
y(t)=e^{2it}(1, \frac{1}{2})+e^{-2it}(6-2i, -2-i)
\end{equation}
Would this be correct, or did I misinterpret the initial conditions?
Thanks
 A: If$$A=\begin{bmatrix}2&8\\-1&-2\end{bmatrix},$$then$$\exp(tA)=\begin{bmatrix}\sin (2 t)+\cos (2 t) & 4 \sin (2 t) \\ -\frac{1}{2} \sin (2 t) & \cos (2 t)-\sin (2 t)\end{bmatrix}.$$So, if $f(t)=\exp(tA).(6,-2)$, you have $f(0)=(6,-2)$ and $f'(t)=A.f(t)$. So, take\begin{align}\bigl(x(t),y(t)\bigr)&=f(t)\\&=\bigl(6 \cos (2 t)-2 \sin (2 t),-\sin (2 t)-2 \cos (2 t)\bigr).\end{align}
A: In your case, the solution writes
$$
\mathbf{x}(t) = 
a_1 e^{\lambda_1 t} \mathbf{v}_1 +
a_2 e^{\lambda_2 t} \mathbf{v}_2
$$
using the eigenvector/eigenvalue of the matrix $\mathbf{A}$
in the relation
$
\dot{\mathbf{x}}(t) = \mathbf{A} \mathbf{x}(t)
$.
Both coefficients $a_1,a_2$ are found using the provided initial conditions
$$
\mathbf{x}(0) = 
a_1 \mathbf{v}_1 +
a_2 \mathbf{v}_2
$$
which is again a 2-by-2 linear system..
UPDATE
With Matlab commands,
A=[2,8;-1,-2];[U,D]=eig(A);U\[6;-2]

I found $a_k = 3.1820 \pm 1.0607i$ and
$\mathbf{v}_1 = [0.9428;  -0.2357 + 0.2357i]$
Thus
$$
\mathbf{x}(t) = 
a_1 e^{2i t} \mathbf{v}_1 +
a_1^* e^{-2i t} \mathbf{v}_1^* = 
2 \mathcal{R}
[
a_1 e^{2i t} \mathbf{v}_1
]
$$
UPDATE 2
Because eigenvalues/eigenvectors are complex conjugate,
it holds
$$
\mathbf{x}(0) = 
a_1 \mathbf{v}_1 +
a_1^* \mathbf{v}_1^*=
2 \mathcal{R}(a_1 \mathbf{v}_1)
$$
$$
\mathbf{A} \mathbf{x}(0) = 
\lambda_1 a_1 \mathbf{v}_1 +
\lambda_2 a_2 \mathbf{v}_2 =
4i 
\mathcal{I}
(a_1 \mathbf{v}_1)
$$
Thus
$$
2 a_1 \mathbf{v}_1=
\mathbf{x}(0)-
i
\frac{\mathbf{A}\mathbf{x}(0)}{2}
$$
From here simplifications occur
\begin{eqnarray*}
\mathbf{x}(t) 
&=& 
\mathcal{R}(2 a_1 \mathbf{v}_1 e^{2i t}) \\
&=&
\mathbf{x}(0) \cos(2t) +
\frac{\mathbf{A}\mathbf{x}(0)}{2} \sin(2t)
\end{eqnarray*}
