What is the degree of $(y')^{-2}+5y'=0 $?

Question

In the video(time stamp present) https://youtu.be/L61hIm_WoC8?t=2944 ,

the differential equation $(y')^{-2}+5y'=0$ is said to have no degree referring that it is not polynomial in derivative(y'), "The teacher says that the equation is not polynomial in derivative so the degree can't be calculated and puts a cross on the right side of the equation."

but I guess multiplying the equation with $(y')^2$, it becomes $1+5(y')^3=0$ which is polynomial in derivative(y'), it has degree 3.

So, which one is correct?

Answer 1

From the links provided by the @user45914123, it can be concluded that the degree for the given differential equation is 3.

Answer 2

The degree of a differential equation is an artificial concept which is of no use. The order of a differential equation with the linearity or non-linearity are essential.

The given differential equation $$ (y')^{-2}+5y'=0$$ is a first order equation which is equivalent to $$1+5(y')^3=0$$

This equation is in term equivalent to $$ y'= \sqrt[3] {(-1/5)} $$ which is a first order linear equation.

As you notice the concept of degree is not essential at all and does not play any role in solving the equation.