Method to solve a system of differential equations I'm studying systems of linear equations. 
I'm now specifically studying systems of linear equations of the 1st order, homogeneous:
$Y' = AY$
$A$ as a constant matrix. 
Now I know there are various methods to solve this systems. My professor talked about one of the method that consists on reducing the system into a single equation. Does anyone knows where can I find notes or explanations about how to apply correctly this method? I'm having trouble finding it on the internet. I find a lot of things about the method that uses eigenvalues/eigenvectors but that's a method that my professor doesn't like... Does anyone know a good text that I can look for? 

If there is a $b$ such that $\{b,Ab,...,A^{n-1}b\}$ is a basis (that is,
if the pair $(A,b)$ is completely controllable), then we
can reduce the system to a single differential equation.
In the above basis, $A$ has the form (controllable canonical form)
\begin{bmatrix}
0 & 1 & \cdots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \cdots & 1 & 0 \\
0 & 0 & \cdots & 0 & 1 \\
-a_n & -a_{n-1} & \cdots & -a_2 & -a_1
\end{bmatrix}
and the system of differential equations looks like
$\dot{x}_1 = x_2,..., \dot{x}_{n-1} = x_n$, $\dot{x}_n = -a_n x_1 - \cdots - a_1 x_n$, and by expanding we get the equation
$x_1^{(n)} + \sum_{k=0}^{n-1} a_{n-k}x_1^{(k)} = 0$, which is a $k$th order differential equation in $x_1$.
Such a $b$ exists iff in the Jordan normal form of $A$, there is exactly one Jordan block of $A$ associated with any eigenvalue.
