Any system of second order ODEs in $m$ variables of the form $$ \frac{d^2{\bf x}}{dt^2} = A{\bf x} + B \frac{d {\bf x}}{dt}$$ can be reduced to a system of first order ODEs in twice as many eigenvalues. The procedure is spelled out in the first video, and follows several steps.

1. Define ${\bf y} = \frac{d{\bf x}}{dt}$.
2. Notice that the second derivative of ${\bf x}$ is the first derivative of ${\bf y}$, so \begin{eqnarray*} \frac{d{\bf x}}{dt} & = & 0 {\bf x} + I{\bf y} \cr \frac{d{\bf y}}{dt} & = & A{\bf x}+ B{\bf y} \end{eqnarray*}
3. Rewrite this as a first-order system $$\frac{d}{dt} \begin{pmatrix} {\bf x} \cr {\bf y} \end{pmatrix} = \begin{pmatrix} 0 & I \cr A & B \end{pmatrix} \begin{pmatrix} {\bf x} \cr {\bf y} \end{pmatrix} $$
4. Solve using the methods of Section 5.2, in terms of the eigenvalues and eigenvectors of $M = \begin{pmatrix} 0 & I \cr A & B \end{pmatrix}.$
5. Notice that if $\begin{pmatrix} {\bf b} \cr {\bf d} \end{pmatrix}$ is an eigenvector of $M$ with eigenvalue $\lambda$, then ${\bf d} = \lambda {\bf b}$, and that ${\bf b}$ satisfies $$ \lambda^2 {\bf b} = A {\bf b} + \lambda B {\bf b}.$$
6. The solutions to the original 2nd-order system are then linear combinations of $e^{\lambda_j t} {\bf b}_j$, where $({\bf b}_j, \lambda_j)$ are solutions to the previous equation.

A similar trick works for 2nd order difference equations of the form $${\bf x}(n) = A {\bf x}(n-2) + B {\bf x}(n-1).$$ We define ${\bf y}(n) = {\bf x}(n-1)$, and get the first order equations $$\begin{pmatrix} {\bf y}(n) \cr {\bf x}(n) \end{pmatrix} = \begin{pmatrix} 0 & I \cr A & B \end{pmatrix} \begin{pmatrix} {\bf y}(n-1) \cr {\bf x}(n-1) \end{pmatrix}, $$ which is again governed by the eigenvalues and eigenvectors of $M$. The only difference is that ${\bf y}$ comes before ${\bf x}$ instead of after. Even so, the general solution is a linear combination of terms $\lambda_j^n {\bf b_j}$, where $({\bf b}_j, \lambda_j)$ are solutions to the equation $$ \lambda^2 {\bf b} = A {\bf b} + \lambda B {\bf b}.$$ In the second video we work this out in general, and apply it to the recursion $$ x(n) = x(n-1) + x(n-2)$$ to get a formula for the Fibonacci numbers.