<s>[INST] Solve $y'=\frac{x+y-2}{x-y}$ ODE 
$$y'=\frac{x+y-2}{x-y}$$

$$\begin{vmatrix} 
  1    & 1\\ 
  1 & -1
\end{vmatrix}\neq0 $$
Substation: $x=u+\alpha$ , $y=v+\beta$ 
$\frac{dy}{dx}=\frac{dv}{du}$
$$\begin{pmatrix} 
  1    & 1\\ 
  1 & -1
\end{pmatrix}\begin{pmatrix} 
  \alpha    \\ 
  \beta 
\end{pmatrix} =\begin{pmatrix} 
  2    \\ 
  0 
\end{pmatrix}\Rightarrow \alpha=1\text{ , }\beta=1 $$
So:
$x=u+1$
$y=v+1$ 
$$y'=\frac{dv}{du}=\frac{u+1+v+1-2}{u+1-(v+1)}$$
$$\frac{dv}{du}=\frac{u+v}{u-v}$$
$$\frac{dv}{du}=\frac{u+v}{u-v}$$
$$\frac{dv}{du}=\frac{1+\frac{v}{u}}{1-\frac{v}{u}}$$
$z=\frac{v}{u}\Rightarrow v=uz$
$$\frac{dv}{du}=z+u\frac{dz}{du}=\frac{1+z}{1-z}$$
How should I continue?
 [/INST] $u\frac{dz}{du}=\frac{1+z}{1-z}-z=\frac{z^2+1}{1-z}$ thus
$$\frac{1-z}{1+z^2}dz=\frac{1}{u}du$$
or
$$\left(\frac{1}{1+z^2}-\frac{z}{1+z^2}\right)dz=\frac{1}{u}du$$
as a result
$$\tan^{-1}(z)-\frac12\ln(1+z^2)=C+\ln(u)$$
</s>