We know that any differential equation (of any order) can be written in the form of a system. Therefore we will be able to solve those equations of order $$n$$ that can be written in a system form that we can solve, a system with constant coefficients.
Let's suppose that we have an ODE of order $$n$$, $$$x^{n)}(t)+a_{n-1}(t) \cdot x^{n-1)}(t)+ \ldots + a_1(t) \cdot x'(t)+a_0(t) \cdot x(t)=f(t)$$$
We define: $$$\begin{array}{rcl} y_1 &=& x \\ y_2 &=& x' \\ y_3 &=& x'' \\ \ldots \\ y_n &=& x^{n-1)} \end{array}$$$ Then $$$\begin{array} {l} y_1'=x'=y_2 \\ y_2'=x''=y_3 \\ \ldots \\ y_n'=x^{n)}=f(t)-a_{n-1}(t) \cdot x^{n-1)}(t)- \ldots -a_1(t)\cdot x'(t)-a_0(t)\cdot x(t)= \\ =f(t)-a_{n-1}(t) \cdot y_n(t)- \ldots -a_1(t)\cdot y_2(t)-a_0(t)\cdot y_1(t)\end{array}$$$ Obtaining a linear system of order 1. As we already said, if it is a system with constant coefficients we will be able to solve it.
Giving the solution will consist in giving the function $$y_1(t)$$, since $$x(t)=y_1(t)$$.
For example, consider the ODE of order 5: $$$x^{5)}+21x^{3)}-2x''-x=2\cos t$$$ We name: $$$\begin{array}{rcl} y_1 &=& x \\ y_2 &=& x' \\ y_3 &=& x'' \\ y_4 &=& x''' \\ y_5 &=& x^{4)} \end{array}$$$
Therefore we obtain the system: $$$\begin{array} {l} y_1'=y_2 \\ y_2'=y_3 \\ y_3'=y_4 \\ y_4'=y_5 \\ y_5'=-21y_4+2y_3+y_1+2 \cos t \end{array}$$$
that we can write in matrix form: $$$\begin{pmatrix} y_1' \\ y_2' \\ y_3' \\ y_4' \\ y_5' \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 2 & -21 & 0 \end{pmatrix}\begin{pmatrix} y_1 \\ y_2 \\ y_3 \\ y_4 \\ y_5 \end{pmatrix}+\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 2\cos t \end{pmatrix}$$$
This method for obtaining a linear system is completely general. Nevertheless, bearing in mind that we can only solve systems with constant coefficients, this method will only be useful to solve ODE's of order $$n$$ with constant coefficients (since with this transformation a triangular system is never obtained).