Homogeneous ODE I am a bit confused with the definition and the solutions ways for homogeneous ODE.
I understand that they are 2 different definitions for homogeneous ODE


*

*The ODE is a function of $y$ and its derivatives such that $F(y,y',y'',...,y^{(n)})=0$ 

*the ODE has the same order of homogeneous such that $M(\lambda x,\lambda y)=\lambda^n M( x, y)$ and $N(\lambda x,\lambda y)=\lambda^n N( x, y)$ which is the same as saying that the ODE can be written in as the function $f(\frac{y}{x})$
If all of the above is correct so if we take $$y(x-y)dx-x^2dy=0$$
We can tell straight away that it is not exact as $M_{y}=x\neq N_{x}=2x$
So we have to find the order of homogeneous:
$M(\lambda x,\lambda y)=\lambda xy-(\lambda y)^2=\lambda xy-\lambda^2 y\neq \lambda^nM(x,y)$
$N(\lambda x,\lambda y)=(\lambda x)^2=\lambda^2x^2=\lambda^2 N(x,y)$
But on the other hand if we divide the ODE by $x^2$ we get:
$$(\frac{y}{x}-\frac{y^2}{x^2})dx-dy=0$$
So it is a non-homogeneous ODE?

setting $$y=xu$$ in your equation then we get with $$y'=u+xu'$$
$$u-u^2-u-xu'=0$$ and you will get $$-\frac{dx}{x}=\frac{du}{u^2}$$ which is easy to solve
