How to find degree of a differential equation.

I looked at some of the classical books in ODE. Most books including Coddington Levinson, Hartman, Chicone do not define the degree of a differential equation. The only book where I found it is Ince. He writes

An ordinary differential equation expresses a relation between an independent variable, a dependent variable and one or more differential coefficients of the dependent with respect to the independent variable.
The order of a differential equation is the order of the highest differential coefficient which is involved. When an equation is polynomial in all the differential coefficients involved, the power to which the highest differential coefficient is raised is known as the degree of the equation. When, in an ordinary or partial differential equation, the dependent variable and its derivatives occur in the first degree only, and not as higher powers or products, the equation is said to be linear. The coefficients of a linear equation are therefore either constants or functions of the independent variable or variables.

And then he gives the following example

$$\left\{1+\left(\frac{dy}{dx}\right)^2\right\}^{1/2}=3\frac{d^2y}{dx^2}$$ is an ordinary equation of the second order which when rationalised by squaring both members is of the second degree.

It is a bit archaic, in particular, by differential coefficients he just means the derivatives of the solution. Anyway, the way I read it is that if a differential equation can be written in the form \begin{align}a_n(x,y(x)) &\left(\frac{dy^n}{dx^n}(x)\right)^{m_n} + a_{n-1}(x,y(x)) \left(\frac{dy^{n-1}}{dx^{n-1}}(x)\right)^{m_{n-1}}\\ &+ \ldots + a_1(x,y(x)) \left(\frac{dy}{dx}(x)\right)^{m_1} = f(x,y(x)),\end{align} where $m_n,\dots, m_1$ are natural numbers and $a_n\ne 0$, then the degree of the ODE is $m_n$.