I have a dynamic system with input $u$ and output $y$ that is described by the following ODE
$$ 10\ddot{y}(t) + \dot{y}(t) = u^{2}(t) $$
that I want to linearize around $u=1$ and determine its transfer function $G(s) = \dfrac{Y(s)}{U(s)}$.
I start with assigning $x_{1} = y$ and $x_{2} = \dot{y}$ which gives
$$ \begin{aligned} \dot{x_{1}} &= \dot{y} = x_{2} \\ \dot{x_{2}} &= \ddot{y} = \frac{u^{2}(t) - x_{2}}{10} \end{aligned} $$
Now to linearize I use that
$$ \Delta\dot{x} = A\Delta x + B\Delta u $$
$$ \Delta y = C\Delta x + D\Delta u $$
$$ \Delta x = x - x_{0},\ \Delta u = u - u_{0} $$
where $A = \frac{\partial f}{\partial x_{i}}$ and $B = \frac{\partial f}{\partial u_{i}}$ where both are evaluated at our operating point $(x_{0}, u_{0})$. C and D are obtained in similar ways but differentiating $g$ (corresponding to the equation for y) and not $f$. So we need to find out operating point. We know that $u_{0} = 1$ and to get $x_{0}$ we can set $\dot{x}=0$ and solve for $x_{10}$ and $x_{20}$. Though this is where I am having problems. Obtaining $x_{10}$ is simple enough, from our assignment $x_{1} = y$ we simply get that $x_{10} = y_{0}$. What is weird is that I am getting conflicting answers for $x_{20}$, see below.
$$ \begin{equation} \begin{split} \dot{x_{1}} &= x_{20} = 0 \\ \dot{x_{2}} &= \frac{u^{2}_{0} - x_{20}}{10} = 0 \implies x_{20} = u_{0}^{2} \end{split} \end{equation} $$
So is the operating point $(x_{10}, x_{20}, u_{0}) = (y_{0}, 0, 1) $ or $(x_{10}, x_{20}, u_{0}) = (y_{0}, u^{2}_{0}, 1) $?
After getting the operating point I would continue and obtain the following matrices
$$ A = \begin{bmatrix} 0 & 1 \\ 0 & -\frac{1}{10} \end{bmatrix}, \qquad B = \begin{pmatrix} 0 \\ \frac{2u\dot{u}}{10} \end{pmatrix}, \qquad C = \begin{bmatrix} 1 & 0 \end{bmatrix}, \qquad D = 0 $$
Evaluating these at the operating point and plugging into the expression for $\Delta \dot{x}$ and $\Delta y$ yields the linearization. Then to obtain the transfer function I can simply use that
$$ G(s) = C(sI - A)^{-1}B + D $$
All doable as long as I have the operating point, but as mentioned previously I am having trouble finding it.
As $u^2\approx u_0^2+2u_0(u-u_0)$ and $10\ddot y_0+\dot y_0 = u_0^2$ we have
$$ 10\ddot y+\dot y = 10\ddot y_0+\dot y_0 +2u_0(u-u_0) $$
or
$$ 10\delta\ddot y+\delta\dot y = 2u_0\delta u $$
with $\delta y = y - y_0, \ \delta u = u - u_0$ and now we can apply the Laplace transform.
NOTE
With null initial conditions we have $y_0 = u_0^2 t$ and $(y_0, u_0)$ is the operating orbit.