Verify the real solution of a linear system of differential equation I'm trying to solve $Y' = AY$ where $A= \left[
  \begin{array}{ c c }
     -2 & 6 \\
     -3 & 4
  \end{array} \right]$
I have found the eigenvalue $1 \pm 3i$ with eigenvector for $1+3i: $
$v =  \left[
\begin{array}{ c c }
     1-i \\
     1 
  \end{array} \right]$
Which seems to be correct by testing $Av = (1+3i)v$ but when I try to write it as a real solution I don't seem to get the right answer.
$$\left[
\begin{array}{ c c }
     1-i \\
     1 
  \end{array} \right](\cos 3t + i\sin 3t) = \left[
\begin{array}{ c c }
     \cos 3t + \sin 3t - i\cos 3t + i\sin 3t \\
     \cos 3t + i\sin 3t 
  \end{array} \right]$$
$$v_1 = \left[
\begin{array}{ c c }
     \cos 3t + \sin 3t \\
     \cos 3t 
  \end{array} \right],\ v_2 = \left[
\begin{array}{ c c }
     \cos 3t + \sin 3t \\
     \sin 3t 
  \end{array} \right]$$
If I now verify by $v_1' = Av_1$ 
$$\left[
\begin{array}{ c c }
     3(\cos 3t - \sin 3t) \\
     -3\sin 3t 
  \end{array} \right] \neq  \left[
\begin{array}{ c c }
     4\cos 3t - 2\sin 3t \\
     \cos 3t - 3\sin 3t 
  \end{array} \right]$$

You are still missing an exponential term, we have:
$e^{\lambda t}v_1 = e^{(1+3i)t}\begin{bmatrix}1-i\\1\end{bmatrix} = e^te^{3it}\begin{bmatrix}1-i\\1\end{bmatrix} = e^t(\cos 3t + i \sin 3t)\begin{bmatrix}1-i\\1\end{bmatrix} = \begin{bmatrix}e^t(\sin 3t + \cos 3t+i (\sin 3 t-\cos 3 t))\\ e^t(\cos 3t + i \sin 3t) \end{bmatrix} $
So, our solution can be written as (because we know that the real and imaginary parts are both independent solutions):
$$Y(t) = c_1 e^t\begin{bmatrix}\sin 3t + \cos 3t\\ \cos 3t \end{bmatrix}+ c_2e^t\begin{bmatrix}\sin 3t - \cos 3t\\ \sin 3t \end{bmatrix}$$
