To understand how a system of ODEs will arise in practical applications, refer to the motivation for eigenvalues.
Given an $n^{th}$ order ODE \begin{align} y^{(n)} = F(t, y, \diffone{y}, \ldots, y^{(n-1)}) \end{align}
we can convert it to a system of ODEs by making the substituitions \begin{align} y_{1} &= y\newline y_{2} &= \diffone{y}\newline &\vdots\newline y_{n} &= y_{(n-1)} \end{align}
This gives the following system of first order equations \begin{align} \diffone{y_{1}} &= y_{2}\newline \diffone{y_{2}} &= y_{3}\newline &\vdots\newline \diffone{y_{n}} &= F(t, y_{1}, y_{2}, \ldots, y_{n-1}) \end{align}
We can test it out on a mass on a spring problem \begin{align} m\difftwo{y} + c\diffone{y} + ky &= 0\newline \implies \difftwo{y} &= -\frac{c}{m}\diffone{y} - \frac{k}{m}y\newline \end{align} Making the substituitions \begin{align} y_{1} &= y\newline \diffone{y_{1}} &= y_{2}\newline \diffone{y_{2}} &= -\frac{c}{m}y_{2} - \frac{k}{m}y_{1}\newline \implies \diffone{\detm{\begin{matrix}y_{1}\newline y_{2} \end{matrix}}} &= \begin{bmatrix}0 &1\newline -\frac{k}{m}y_{1} &-\frac{c}{m}y_{2}\end{bmatrix}\begin{bmatrix}y_{1}\newline y_{2}\end{bmatrix} \end{align} We can solve for this system by finding the eigenvalues of the matrix. The characteristic equation is \begin{align} (-\lambda)(-\frac{c}{m}-\lambda) + \frac{k}{m} &= 0\newline \implies \lambda^{2} +\frac{c}{m}\lambda + \frac{k}{m} &= 0 \end{align}