How do I find poles and zeros from an ODE? Assume that you having an ODE.
$$\dot x = x^3y x^2 + x + y + 10$$
How can I find the poles and zeroes here?
If I was using a transfer function,
$$G(s) = \frac{s^2 + 1}{s^2 + 0.1s + 50}$$
then the poles would be the roots of $0 = s^2 + 0.1s + 50$ and the zeros would be the roots of $0 = s^2 + 1$
But in this case, I don't have a transfer function. Just a regular ODE.
 A: Poles and zeros are a concept that is defined for transfer functions. Transfer functions are a concept that is defined for linear, time invariant systems.
Your system however is nonlinear, so the concept of poles/zeros is not defined for such systems; so it doesn't make sense to ask for the poles/zeros of such a system.
You can however linearize the system, as that gives you a linear system which has poles and zeros that you can analyze. Since you say the $y$ is an input and $x$ an output I redefine the input to $u$ and the output to $y$ which is more standard:
\begin{align}
\dot{x}&=x+10+x^5u+u\\
y&=x
\end{align}
First note that this system does not have a unique equilibrium. So for linearization, you have to choose which equilibrium point you want to analyze. For example, say you are interested in the equilibrium $x=0,u=-10$. The linearization matrices are:
$$
\begin{align}
A&=5 u x^4 + 1\\
B&=x^5 + 1\\
C&=1\\
D&=0
\end{align}
$$
So at $x=0,u=-10$ you get:
$$
\begin{align}
A&=1\\
B&=1\\
C&=1\\
D&=0
\end{align}
$$
This can be used to compute the transfer function
$$
G(s)=C(sI-A)^{-1}B+D
$$
($I$ is the identity matrix). So if you plug everything in you get
$$
G(s)=\frac{1}{s-1}
$$
So at the equilibrium $x=0,u=-10$, the linearized system of your nonlinear system has no zeros and one pole at $s=1$. Since the pole has positive real part, this equilibrium is unstable.
Of course, the linearization is an approximation that is only valid in a neighborhood of the equilibrium point. At other equilibrium points the system might have a very different behaviour.
