Stopień ODE [nie] [duplikat]

Question

Stopień równania różniczkowego, gdy jest to równanie wielomianowe w pochodnych, jest najwyższą potęgą pochodnej najwyższego rzędu.

Przykłady

ja) $\frac{d^3y}{dx^3}-\frac{dy}{dx}+xy^2=0$ Stopień to 1

ii) $(\frac{dy}{dx})^2+\frac{dy}{dx}+x\sin(y)=0$ Stopień to 2

iii) $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\cos\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=0$ Stopień nie jest określony

Dlaczego stopień $$\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+\cos\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=0$$ not defined, since it is still a polynomial in $y''$?

Edit:

Considering the fact that different sources provide different definitions, I look for a reliable and complete definition of what a degree is, and why it is/ isnt defined in this case

Answer 1

Wikipedia states the definition of degree of ODE as: 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.

According to you , polynomial should be formed in highest order derivative, which is not the case. The polynomial should be in all of it's derivatives, and hence the degree of the DE $y''+\cos(y')=0$ is not defined.