This question is from a feedback control lecture. I do not understand how to compute this, please explain interactively.
Given this differential equation,
$\frac{d^3z}{dt^3} + \frac{5d^2z}{dt} + \frac{3dz}{dt} + 2z = u$
In state-space representation, put the system in the form x' = Ax + Bu
The given answer to this question was
\begin{equation}x' = \begin{bmatrix}0&1&0\\0&0&1\\ -2 & -3 & -5\end{bmatrix}x + \begin{bmatrix}0\\0\\ 1\end{bmatrix}u\end{equation}
The idea is to introduce new variables in order to convert the third order equation into an equation of order one in higher dimensions.
In this case you make $dz=y$ and $dy=x$, then $d^2z=dy=x$ and $d^3z=dx$ so your equation becomes the one shown in the solution for a three variable set up considering your vector $X=\left( \begin{array}{c} x\\ y\\ z \end{array} \right)$, $X'=\left( \begin{array}{c} dx\\ dy\\ dz \end{array} \right)=\left( \begin{array}{c} d^3z\\ x\\ y \end{array} \right)=\left( \begin{array}{c} -5x-3y-2z+u\\ x\\ y \end{array} \right)$ which is the same up to a reordering of the rows.