Transform the following equation system of order $$2$$:
$$\displaystyle x''+\frac{b}{m} \cdot x'+\frac{k}{m}\cdot x =0$$
Development:
We have a linear ODE of order 2. We define two new variables: $$$y_1=x \ \ \ \ y_2=x'$$$ So the ODE can be written as follows: $$$\left. \begin {array} {l} y_1'=y_2 \\ y_2'=-\dfrac{b}{m}\cdot y_2-\dfrac{k}{m}\cdot y_1 \end{array}\right\} \Rightarrow \begin{pmatrix} y_1' \\ y_2' \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ -\dfrac{k}{m} & -\dfrac{b}{m} \end{pmatrix}\cdot\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$$$ If we wanted to solve the ODE, it would be necessary to solve the system and give as a solution $$x(t)=y_1(t)$$.
Solution:
$$\begin{pmatrix} y_1' \\ y_2' \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ -\dfrac{k}{m} & -\dfrac{b}{m} \end{pmatrix}\cdot\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}$$