How to find degree of a differential equation. I have a differential equation, $$e^{\large y^\prime} = x + x^3 + x^5 + y,$$ I need to find the degree of this equation. 
Using Wikipedia definition,

In mathematics, the degree of a differential equation is the power of its highest derivative, after the equation has been made rational and integral in all of its derivatives.

I would say that the degree is one because if I take $\log $ on both sides I get $${y^\prime} = \log(x + x^3 + x^5 + y).$$
My teacher says that the degree is not defined because this DE cannot be represented as sum of polynomials in derivatives of $y$. When I asked, what if we take $\log$ on both sides, he says that we are not allowed to perform any operations on the DE, that will change DE of which we have to find the degree.
This contradicts the definition by Wikipedia.
Who is correct? What is the degree of this DE, $1$ or not defined?
 A: 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 $m_n$.
