I want to find the transfer function for a system described by this non-linear state-space model: \begin{equation} X(t) = \begin{bmatrix} y(t)\\ v(t)\\ \varphi (t)\\ \omega (t)\\ T_d(t) \end{bmatrix} \Rightarrow \dot X(t) = \begin{bmatrix} \dot y\\ \dot v\\ \dot \varphi\\ \dot \omega\\ \dot T_d \end{bmatrix} = \begin{bmatrix} v\\ y\omega ^2 - g\\ \omega\\ \frac{T_d}{my^2} - \frac{g}{y} - \frac{2\dot y\omega}{y}\\ \frac{1}{\tau}(k_m u(t) - T_d(t)) \end{bmatrix} \end{equation} \begin{equation} y(t) = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \end{bmatrix} x(t) \end{equation} The operating point for equilibtium state is given by:
$(\varphi _0 = 0)$. \begin{equation} T_{d}0=mgy_0 \end{equation}
This is how I obtain the transfer function: \begin{equation} s\Delta X_{1}(s) = \Delta X_{2}(s) \end{equation} \begin{equation} s\Delta X_{2}(s) = \Delta X_{1}(s)\Delta X_{4}(s) - g) \end{equation} \begin{equation} s\Delta X_{3}(s) = \Delta X_{4} \end{equation} \begin{equation} s\Delta X_{4}(s) = \frac{\Delta X_{5}(s)}{m\Delta X_{1}^2} - g\frac{1}{\Delta X_{1}(s)} - \frac{2\Delta X_{1}(s)\Delta X_{4}(s)}{\Delta X_{1}(s)} \end{equation} \begin{equation} s\Delta X_{5}=\frac{1}{\tau}(kmu(t) - \Delta X_{5}(s)) \end{equation} \begin{equation} \Delta Y(s) = \Delta X_{1}(s) \end{equation} \begin{equation} \Delta X_{5}(s) = \frac{kmU(s)}{s\tau + 1} \end{equation} \begin{equation} \frac{K_{m}}{T_{d}(1 + \tau s)} \end{equation}
Your system has no equilibrium point, so a transfer function from a linearization might not represent the actual system behavior even when the system is arbitrarily close to your point of linearization.
Note that you divide by $y$, so we need $y\neq 0$. However, this means that $\dot{v}=0$ only if $y=g/\omega^2$. Since you divide by $\omega^2$, it follows that $\omega\neq 0$ is necessary. However, $\dot{\varphi}=0$ if and only if $\omega=0$, a contradiction. So there is no operating point that satisfies the requirement of an equilibrium, namely that $\dot{x}=0$.
Note that this argument assumes $g\neq 0$. I guess $g$ stands for gravity of earth, so this assumption should be justified. If this is not true and in fact $g=0$ then you have an equilbrium at $v=0,\omega=0,T_d=0,u=0,\varphi\in R$ and $y\in R\setminus\{0\}$.
However, the transfer function in this case is $G(s)=0$ (independently of the choice of $\varphi$ and $y$), so not very useful either.