Find an integrating factor such that $y'=\frac{1-x+y}{x-y}$ is exact Yet another question of this sort, and hopefully the last.
In the previous question I posted, we were lucky enough and the integrating factor was a function of only one variable, the ansatz $\mu_y=0$ was successful.
In this question, it won't help us.
After simplifying $y'=\frac{1-x+y}{x-y}$ we get $(x-y-1)dx+(x-y)dy=0$. It is not exact.
We need to find $\mu (x,y)$ such that $\mu(x-y-1)dx+\mu(x-y)dy=0$ is exact. 
This means that $\mu_y(x-y-1)-\mu=\mu_x(x-y)+\mu$
And no matter what ansatz we use (either $\mu_x=0$ or $\mu_y=0$), $\mu$ seems to depend on both variables. How unlucky.
How would we find $\mu$ in this case?

Using integrating factor to solve this ODE isn't the simplest method. Nevertheless, since that is the requested method, an integrating factor is found below :

From the above implicit solution, it is possible to express $y(x)$ on explicit form, thanks to the Lambert W function.
