How to calculate A and B matricies in Control Theory for my system given the ODE describing the system? I have an ODE describing my system like so:
$\ddot{x} = -\Gamma_0\dot{x} - (\Omega_0 + u)x$
Where u is my feedback control actuator and $\Gamma_0$ and $\Omega_0$ are constants.
I want to find the $\mathbf{A}$ and $\mathbf{B}$ matrix for this system such that I have
$\dot{\vec{x}} = \mathbf{A}\vec{x} + \mathbf{B}\vec{u}$
Where $\vec{x}$ is my state vector 
$\vec{x} = \begin{bmatrix} 
    x \\
    \dot{x} \\
\end{bmatrix}$
and $\vec{u} = u$
I am unsure how to proceed to get $\mathbf{A}$ and $\mathbf{B}$.
I understand that if my dynamics were non-linear I would find the fixed points of the non-linear dynamics and then linearise about that point by calculating the Jacobian of the dynamics and using this as my A matrix however I am not sure how to proceed as the actuator introduces the non-linearity.
Written as 2 first order differential equations I get
$$\begin{bmatrix} 
    \dot{x} \\
    \ddot{x} \\
\end{bmatrix} = 
\begin{bmatrix} 
    0 & 1 \\
    -(\Omega_0 + u) & -\Gamma_0 \\
\end{bmatrix}
\begin{bmatrix} 
    x \\
    \dot{x} \\
\end{bmatrix} 
$$
and I can split these up into the following 2 terms. 
$$\begin{bmatrix} 
    \dot{x} \\
    \ddot{x} \\
\end{bmatrix} = 
\begin{bmatrix} 
    0 & 1 \\
    -(\Omega_0) & -\Gamma_0 \\
\end{bmatrix}
\begin{bmatrix} 
    x \\
    \dot{x} \\
\end{bmatrix} 
+ 
\begin{bmatrix} 
    0\\
    -x \\
\end{bmatrix}
u
$$
But it is my understanding that to implement control using closed-loop feedback control using $u=-\mathbf{K}\vec{x}$ the $\mathbf{A}$ and $\mathbf{B}$ must be constant matrices.

Denote $\bar x=(x,\dot x)$; the system is
$$
\dot{\bar x} = f(\dot{\bar x}),\quad 
f(\dot{\bar x})=
\begin{bmatrix} 
    \dot x\\
    -(\Omega_0 + u)x -\Gamma_0\dot x \\
\end{bmatrix}
$$
$$
A=\left.\frac{\partial f}{\partial x}\right|_{\bar x=0,u=0}=
\left.\left(\begin{array}{cc}
0&1\\
-(\Omega_0 + u)&-\Gamma_0
\end{array}\right)\right|_{\bar x=0,u=0}=
\left(\begin{array}{cc}
0&1\\
-\Omega_0&-\Gamma_0
\end{array}\right)
$$
$$
B=\left.\frac{\partial f}{\partial u}\right|_{\bar x=0,u=0}=
\left.\left(\begin{array}{c}
0\\
-x
\end{array}\right)\right|_{\bar x=0,u=0}=
\left(\begin{array}{c}
0\\
0
\end{array}\right)
$$
The controllability matrix is equal to
$$
U=(B,AB)=\left(\begin{array}{cc}
0&0\\
0&0
\end{array}\right)
$$
