I have this system of ODEs and I'm trying to get a linearized version of it around the "operating point" $\overline{x}_1 = 1$
$$ \left\{\begin{matrix} \ddot{x_1}(t)+2\dot{x_1}(t)+2x_1^2(t)-2\dot{x_2}(t)=0 \\ 2\ddot{x_2}(t)+2\dot{x_2}(t)-2\dot{x_1}(t)=f(t) \end{matrix}\right. $$
So I define a small perturbation $\delta x_1$, $\delta x_2$ and $\delta f$ around the operating point $\overline{x}_1$, $\overline{x}_2$ and $\overline{f}$
$$\delta x_1 = x_1 - \overline{x}_1 \Rightarrow \dot{x_1} = \dot{\delta x_1} \Rightarrow \ddot{x_1} = \ddot{\delta x_1}$$
$$\delta x_2 = x_2 - \overline{x}_2 \Rightarrow \dot{x_2} = \dot{\delta x_2} \Rightarrow \ddot{x_2} = \ddot{\delta x_2}$$
$$\delta f = f - \overline{f} $$
I use Taylor polynomial to linearize $x_1^2(t)$ around $\overline{x}_1=1$ as
$$ x_1^2 \approx \overline{x}_1^2 + 2\overline{x}_1 \delta x_1 = 1 + 2\delta x_1$$
I replace all in the original equations:
$$\left\{\begin{matrix}\delta\ddot{x_1}(t)+2\delta\dot{x_1}(t)+2\left [1+2\delta x_1(t) \right ] - 2 \delta \dot{x_2}(t)=0 \\ 2\delta \ddot{x_2}(t)+2\delta \dot{x_2}(t)-2\delta \dot{x_1}(t)=\overline{f}+\delta f(t) \end{matrix}\right.$$
This system is "linear", but not homogeneous, because it has constant terms $2$ and $\overline{f}$. In fact, through force balance we get that $\overline{f}=2$, so the constant terms should mathematically cancel out somehow.
How can I get rid of this constant terms? Is there another (better) way to linearize this system of ODEs around $\overline{x}_1=1$
By the way, I got this systems of ODEs from this physical system:
When linearising a non-linear system of the form $\dot{x} = g(x,f)$ at an operating point $\bar{x}$ and $\bar{f}$ requires that $g(\bar{x},\bar{f})=0$. Since $\bar{x}_1$ is given and $g(x,f)$ is not a function of $x_2$, then $g(\bar{x},\bar{f})=0$ only has a solution when $x_2$ is omitted from the state space vector, so $x$ only contains $x_1$, $\dot{x}_1$ and $\dot{x}_2$ and no $x_2$. So $\bar{\dot{x}}_2$ can then be a non-zero constant, which can be chosen such that $g(\bar{x},\bar{f})=0$ can be satisfied.