State space and linearization I have a question about state space representation. How can I represent an equation in which I have only the second and first derivatives? For example

where $u$ is the control input.
If I put $x_1=x$ and $x_2=\dot{x}$ I will not have $x_1$ in my state space representation, and when finding equilibrium points or Jacobian to check controllability, I will obtain zero in the partial derivatives corresponding to $x_1$. Is there a way to overcome this with a more suitable state representation?
 I'm not sure how to solve this problem as I have to linearize the system around an operating point using SS. Thanks in advance.

By selecting $x_1=x$ and $x_2=\dot{x}$ you can write
$$\begin{align*}
\dot{x}_1 &= x_2 \\
\dot{x}_2 &= k_1 x_1 (1 - |x_1|) + k_2 u
\end{align*}$$
Now the fixed points are $(0, 0), (1, 0), (-1, 0)$. The Jacobian around $(1, 0)$ (as an example) is
$$ \begin{bmatrix} 0 & 1 \\ k_1(1 - 2 x_1) & 0 \end{bmatrix} |_{x_1=1} = \begin{bmatrix} 0 & 1 \\ -k_1 & 0 \end{bmatrix} $$
