I am trying to linearize the following affine relationship where $x \in \mathbb{R}^n, u \in \mathbb{R}^m$, and $ c \in \mathbb{R}^n$:
$$\dot{x}(t)=f(x,u,t)=A(t)x(t)+B(t)u(t)+c(t)$$
Defining the equilibrium point, $\bar{x}$ to be where $\dot{x}=0$, I find that this equilibrium point must vary with time:
$$\bar{x}(t)=-A(t)^{-1}\left[B(t)\bar{u}(t)+c(t)\right]$$
Problem: when I try to linearise around the above fixed point $(\bar{x},\bar{u}$) using the perturbation variables $\delta x=x-\bar{x}$ and $\delta u=u-\bar{u}$, in an attempt to re-write the system as:
$$\delta\dot{x}=A(t)\delta{x}+B(t)\delta{u}$$
I get stuck since to me it seems that $\delta\dot{x}=\dot{x}-\dot{\bar{x}}$ with $\dot{\bar{x}}\ne0$.
Doesn't this contradict the fact that $f(\bar{x},\bar{u},t)=\dot{\bar{x}}=0 \Longrightarrow \dot{\bar{x}}=0$?
Basically, I don't see how to linearise this time varying system around an equilibrium point.