homogeneous differential equations understanding example $(y-x)y'=2x$

Question

I'm reading an example on homogeneous differential equations which goes like this:

$$ (y-x)y'=2x $$

Understand that in those case of scenarios we let $$y=y(x)=xz(x)$$ After we differentiate we get $$y'=z+xz'=\frac{2}{z-1}$$

I really don't understand why $z+xz'=\frac{2}{z-1}$. How did we get that?

Thanks in advance!

Answer 1

It's actually really simple. $$ z+xz'=y'=\frac{2x}{y-x} = \frac{2}{\frac{y}{x}-1} = \frac{2}{z-1} $$ Because for non-zero $x$ $$ y = xz \iff z=\frac{y}{x} $$

Answer 2

$$(y-x)y'=2x \implies y'= \frac {2x}{y-x} = \frac {2}{y/x-1}=\frac {2}{z-1}$$

$$ y= xz \implies y' = xz' + z$$

$$ xz' + z =\frac {2}{z-1}$$

$$xz' =\frac {2}{z-1}- z$$

which is separable.

Answer 3

Assume $x\neq 0$, which excludes $y= x\iff z=1$. Then you have $y'=2x/(y-x)$ and $y=xz$, then$$ y'= 2x/(xz-x)=2/(z-1), $$ and you can conclude with the identity $y'=z+xz'$.