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.
 A: 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
$$
U=(B,AB)=\left(\begin{array}{cc}
0&0\\
0&0
\end{array}\right)
$$
A: An alternative approach, which does not require you to go to the state space formulation is as follows. First, rewrite the ODE such that everything is on one side of the equation. Then we can rewrite the ODE as:
$$\Phi(\ddot{x},\dot{x},x,u)=\ddot{x} +\Gamma_0\dot{x} + (\Omega_0 + u)x=0.$$
In order to linearize the ODE we need to evaluate the Taylor series of $\Phi$. It is given by:
$$\Phi(\ddot{x},\dot{x},x,u)\approx\left.\Phi(\ddot{x},\dot{x},x,u)\right|_{\text{eq}}+\left.\dfrac{\partial \Phi}{\partial \ddot{x}}\right|_{\text{eq}}\left[\ddot{x}-\left.\ddot{x}\right|_{\text{eq}} \right]+\left.\dfrac{\partial \Phi}{\partial \dot{x}}\right|_{\text{eq}}\left[\dot{x}-\left.\dot{x}\right|_{\text{eq}} \right]$$
$$+\left.\dfrac{\partial \Phi}{\partial x}\right|_{\text{eq}}\left[x-\left.y\right|_{\text{eq}} \right]+\left.\dfrac{\partial \Phi}{\partial u}\right|_{\text{eq}}\left[u-\left.u\right|_{\text{eq}} \right]=0.$$
