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Below I've posted a number of ODE whose degrees are to be established. Please post your sincere opinions.

Definitions follow:

Order : The order of a differential equation is the order of the highest order derivative appearing in the given equation. It is always defined and not equal to 0.

Degree : To find out the degree, express the differential equation as a polynomial in differential coefficients ($y'=\frac{dy}{dx}, y''=\frac{d^2y}{dx^2}, \ldots$) and the dependent variable $y$. Then, the power of the highest order differential coefficient will be the degree of the ODE. It may or may not exist.


1. $\log(1+y'') = y'$

$\Rightarrow 1+y''=e^{y'} = 1 + y' + \frac12(y')^{2} \ldots \Rightarrow$ Order = 2, Degree = 1.

2. $\log(1+y'') = y'' $

$\Rightarrow 1+y''=e^{y''} = 1 + y'' + \frac12(y'')^2\ldots \Rightarrow 0= \frac12(y'')^2\ldots \Rightarrow$ Order = 2, Degree: undefined.

3. $\log(1+y'') = y''' $

$\Rightarrow 1+y'' = e^{y'''} = 1 + (y''') + \frac12(y''')^2 \ldots \Rightarrow $Order=3, Degree: undefined.

4. $1 + y'' = \log(y') $

$\Rightarrow 1 + y'' = \log(1+(y' - 1)) = (y' - 1) - \frac12(y' - 1)^2 + \frac13(y' - 1)^3 \ldots \Rightarrow $ Order: 2, Degree: 1.

5. $y = \sin(y')$

In the final case, we do not know where $y$ lies so we can not take $\arcsin$ on both sides and expand $\arcsin(y')$. Order: 1, Degree: undefined.

Are these solved correctly?

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  • $\begingroup$ What is the question though? $\endgroup$ – AlexanderJ93 Sep 17 '18 at 22:11
  • $\begingroup$ Are these solved correctly? $\endgroup$ – So Lo Sep 19 '18 at 20:15
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The concept of "degree" here does not seem to be well defined. Following the instructions given, both (1) and (3) could have either degree 1 or undefined degree, depending on if the $\log$ is inverted first before expanding. The problem is that there is not a unique polynomial representation of an equation. As a simple example, consider $y'=y$. We can raise both sides of this equation to any odd power, and the solution over the reals will not change, i.e. $y'^{17} = y^{17} \Rightarrow y = c\exp(x)$. A solution to this would be to take a minimal degree. In the simple example and in both (1) and (3), the degree would be 1. In fact, every equation of finite odd degree would have degree 1, since we can take odd roots without changing the equation (again, over reals), and expand the root using it's Taylor series.

Indeed, it seems that polynomials aren't quite as natural over the real numbers as they are over the complex numbers. So, another solution would be to consider the degree over the domain of complex numbers. However, in the complex domain, none of the common nonlinear functions are invertible ($\exp, \log,$ trig, powers, roots). This means that anytime the highest derivative appears inside a function, the degree of that function will be the degree of the equation. Thus, if that function is not a polynomial (of finite degree) then it will not have a degree, i.e. undefined.

In reality, the degree of an ODE has little to no bearing on the difficulty or the methods used to solve, so it's not a very meaningful class. We don't often discuss the degree of general ODEs or PDEs in this way. This is evidenced by the fact the Wikipedia entry is basically empty, and cites a few sources which have very little to say on the topic. If you think you have discovered some reason why it may be interesting to look at the degree of a general ODE, then you may want to choose the definition which best fits your application, being especially careful that it is a consistent definition.

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