Homogeneous ODE

Question

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

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

2. 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?

Answer 1

Your DE is not homogenous as it is stated - whichever definition you apply. You checked the second definition, and for the first one you get $y'=\frac{y(x-y)}{x^2}$, and the right hand side is obviously not homogenous with respect to $y$.

However, a change of variables can make your DE homogenous, can you spot it?

Answer 2

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

Answer 3

The equation you have there is actually homogeneous of degree $2$, you made a slight mistake when testing for $M(x,y)$ from the definition of homogeneous of degree wherever there is $x$ or $y$ you multiply by $\lambda $ for each $x$ and $y$ therefore we shall have $(\lambda x) (\lambda y) - (\lambda y)^2 = \lambda^2(x y - y^2)$. Moreover there is a difference between homogeneous of degree ODE and homogeneous linear ODE.