I have been given the following ODE: $$y'=\frac{1}{y-x}$$
I verify that it is homogeneous by doing: $$y'=\frac{1}{\lambda y-\lambda x}$$ $$y'= \frac{1}{\lambda (y-x)}$$ $$y'=\lambda ^{-1} \frac{1}{y-x} $$
Therefore, it is a homogenous equation of degree -1.
At this point, I would do the following substitution: $y=ux$, $y'=u'x+u$.
$$u'x+u=\frac{1}{ux-x}$$ $$u'x+u=\frac{1}{x(u-1)}$$ $$u'x=\frac{1}{x(u-1)}-u$$ $$u'x=\frac{1-ux(u-1)}{x(u-1)}$$ $$u'=\frac{1-ux(u-1)}{x^2(u-1)}$$
I do not know how to continue from here as in all the homogeneous ODE I have seen so far, I was able to separate $x$ and $u(x)$.
Any help would be appreciated.
I am a bit inexperienced at this matter, but here goes my idea:
This doesn't look like an homogeneous equation since: \begin{equation*} f(x,y) \neq f(\lambda x,\lambda y) \end{equation*} So my suggestion is the following: \begin{equation*} \frac{dy}{dx} = \frac{1}{y-x} \Leftrightarrow \frac{dx}{dy} = y-x \Leftrightarrow \frac{dx}{dy} = (-1)x+y \end{equation*}
Which is now a linear equation, solvable with easier methods.