Find a real-valued vector solution to a system of differential equations Given: 
$$\vec x'(t) = \begin{bmatrix}
    4&-1\\
13&0
    \end{bmatrix} \vec x(t) $$
Evaluating to find eigenvalues:
$$ (4-\lambda)(-\lambda)+13=0 $$$$ (\lambda-2)^2=-9$$ $$\lambda=2\pm3i$$
Finding the eigenvector for the eigenvalue $\lambda = 2+3i$:
$$ \left[
\begin{array}{cc|c}
  2-3i&-1&0\\
  13&-2-3i&0
\end{array}
\right] $$
Since, for a 2x2, the rows must be complex scalar multiples of each other (here: R2 = (2 + 3i)R1 ), then picking row 2:
\begin{align*} \begin{cases} v_1=\frac{2+3i}{13}v_2 \\ v_2=free \end{cases} \end{align*}
$$\vec v= v_2 \begin{bmatrix}
    \frac{2+3i}{13}\\
1
    \end{bmatrix} = v_2\begin{bmatrix}
    2+3i\\
13
    \end{bmatrix}$$
The solution takes the form of:
$$e^{\lambda_1t}\vec v_1 = e^{(2+3i)t}\begin{bmatrix}
    2+3i\\
13
    \end{bmatrix}
$$
However, this is not a real-valued vector solution, since the eigenvector has an imaginary part. How should I proceed?

Your matrix is actually similar to one of the form $\begin{bmatrix}
    2&-3\\
3&2
    \end{bmatrix}$ with transition matrix $\begin{bmatrix}
    2&3\\
13&0
    \end{bmatrix}$ given respectively by the eigenvalues' real and imaginary parts and the transition is given (in columns) by real and imaginary parts of the first eigenvector.
The solution to the equation will then be given by $\vec x(t)=e^{2t}\begin{bmatrix}
    cos(3t)& sin(3t)\\
-sin(3t)&cos(3t)
    \end{bmatrix}.\vec x(0)$
To give a proof of this fact, apply Taylor expansion and definition of exponential of a matrix. 
Regards
