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?

  • $\begingroup$ "has been made rational and integral in all of its derivates" ; I have difficulties to understand this ... $\endgroup$ – Peter Sep 19 '17 at 17:26
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    $\begingroup$ @Peter I think it means to write the DE in form $a y_n^m + a_1 y_{n-1}^{m_1} + ... + a_{n}y^m_1 = f(x,y)$ where $m$ is an integer. $\endgroup$ – A---B Sep 19 '17 at 17:41
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    $\begingroup$ @Peter I made a mistake in my previous comment. it should be $m,m_1, ..., m_n$ all are integer. $\endgroup$ – A---B Sep 19 '17 at 17:52
  • $\begingroup$ Even with the log, now you have encased the zeroth order derivative inside a logarithm...this isn't rational or integral in ALL derivatives of $y$. You might decide to replace the exponential with its taylor series expansion....but now you have an infinite polynomial in $y'$ with no highest degree. $\endgroup$ – Triatticus Sep 22 '17 at 4:06
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    $\begingroup$ I don't think this is well-defined: $y'=y$ is degree one, but $(y')^3 = y^3$ encodes exactly the same differential equation, but is degree three. Either you don't allow any manipulation to the equation and define the degree of this very equation in its current form (in your case, it wouldn't be defined) or, if you allow manipulation, you have to state which manipulations you allow in order to make the definition well-defined. $\endgroup$ – Stefan Hante Sep 26 '17 at 7:48

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 $m_n$.

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    $\begingroup$ So, if I had $e^{y^\prime} = (x^2 + x + 1)2^{y}$ then I take $\log$ and get $y^\prime = \log(x^2 + x + 1) + y \log(2)$. So the degree is defined and is $1$ right ? $\endgroup$ – A---B Sep 27 '17 at 7:03
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    $\begingroup$ yes you are right! The degree is one. The log involves only y and not derivatives of y, so it is a polynomial in the derivatives. I removed the last sentence in my answer. $\endgroup$ – Gio67 Sep 27 '17 at 12:22
  • $\begingroup$ The example above is 1st order ODE in terms of the slope $\theta = \frac{{\rm d}y}{{\rm d}x}$ $\endgroup$ – ja72 Sep 27 '17 at 13:28

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