$y'=\frac{x-y+2}{x+y-1}$ First, I substitute $x=\xi + h$, $y=\eta+k$. The equation becomes $$y'=\frac{\xi+h-(\eta+k)+2}{\xi+h+\eta+k-1}$$
and this becomes homogenous if we let $h=-\frac{1}{2}$ and $k=\frac{3}{2}$ ( We solve the constant terms such that they become zero ). I think it's correct this far, but what next? At this point we'd have $$\eta'=\frac{\xi-\eta}{\xi+\eta}$$
which is homogenous, and I tried substituting $\eta = u\xi$. This would lead to the equation $$u'\xi+u=\frac{1-u}{1+u}$$
I reduce $u$ from both sides. This will give $$u'\xi=\frac{1-u}{1+u}-u=\frac{1-2u-u^2}{1+u}$$
or $$\frac{du}{d\xi}\xi=\frac{1-2u-u^2}{1+u}$$
inverting this gives the relation $$\frac{1}{\xi}d\xi= \frac{1+u}{1-2u-u^2}du$$
So there has been some mistake made. The correct answer does not involve logarithms, and $\ln(\xi)$ will inevitably be a part of the solution. Can somebody spot my mistake?

The integral on the RHS also involves logarithm. When you take antilogs on both sides after integration, you will be free of any logarithms

Another method:
$$y'=\frac{x-y+2}{x+y-1}$$
Cross multiply:
$$xdy+ydy-dy=xdx-ydx+2dx$$
$$d(xy)+ydy-dy=xdx+2dx$$
