How do you solve a $n$-th order system of differential equations in the form of $\mathbf x^{(n)}=A\mathbf x$? 
How do you solve a $n$-th order system of differential equations.

I know how to solve first order equations like $\mathbf{x}'=A\mathbf{x}$ but how would I solve $$\mathbf x^{(n)}=A\mathbf x$$ Could you maybe show me how it is done or link me some sources where I can look this up. I have looked at Paul's notes, Mathematical Methods for Physicsists and Engineers and Wikipedia but they seem to all focus on first order equations.
Edit: Winther demonstrated a standard procedure of reducing nth order systems to first order systems in the comments. I am looking for sources that explain this in detail.
 A: Given an $n$'th order ODE
$${\bf y}^{(n)} = A{\bf y}$$
where ${\bf y}$ is a vector in $\mathbb{R}^k$ then we can reduce the order of the equation by introducing auxiliary variables ${\bf y_i} = {\bf y}^{(i-1)}$ for $i=1,2,3,\ldots,n$. With these variables the ODE can be written
$$\matrix{{\bf y_n}' &=& A{\bf y_1}\\
{\bf y_{n-1}}' &=& {\bf y_n}\\
&\ldots&\\
{\bf y_{1}}' &=& {\bf y_2}}$$
which is a closed system of coupled first order equations. The price to pay for this reduction in order is that the $k$ equations we started with have turned into $nk$ equations.
This system can also be written on matrix form ${\bf Y'} = B{\bf Y}$ by working with the vector
$${\bf Y} = \pmatrix{({\bf y_1})^1\\\ldots\\({\bf y_i})^j\\\ldots\\({\bf y_n})^k}$$
The matrix $B$ will contain $A$ in the lower-left corner and the rest of the component are $0$ except for $B_{j,k+j} = 1$ for $j=1,2,\ldots,(n-1)k$.
The general formulation of the method is a bit messy with a lot of vectors and components, but it's not very complicated in practice. As long as you remember the main method here then you will be able to use it for any equation of interest (I would not bother remembering the general formula for $B$ just focus on remembering the steps needed to reduce it to that form).

To show an application we can try to apply this method to your system
$${\bf y''} = A{\bf y}$$
where ${\bf y} = \pmatrix{y^1\\y^2}$ and $A = \pmatrix{13/3 & 20i/3\\20i/3 & -37/3}$. We introduce ${\bf y}_1 = {\bf y}$ and ${\bf y}_2 = {\bf y'}$ so that
$$\pmatrix{{\bf y_1'}\\{\bf y_2'}} = \pmatrix{{\bf y_2}\\A{\bf y_1}}$$ 
which can be written
$${\bf Y'} = \pmatrix{0&0&1&0\\0&0&0&1\\13/3 & 20i/3&0&0\\20i/3 & -37/3&0&0}{\bf Y}$$
where $Y_1 = {\bf y^1_1} = {\bf y^1}$, $Y_2 = {\bf y^2_1} = {\bf y^2}$, $Y_3 = {\bf y^1_2}$, $Y_4 = {\bf y^2_2}$ (here a super-script denotes the component number).
