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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?

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  • $\begingroup$ try it with $\frac{-x+y}{-2x+2y+1}$ $\endgroup$ – Dr. Sonnhard Graubner Nov 23 '14 at 19:29
  • $\begingroup$ How did you reach this result? $\endgroup$ – Oria Gruber Nov 24 '14 at 0:11
  • $\begingroup$ @Gruber, i don't know if it help in finding the integrating factor, but you can solve the de with a substitution for the dependent variable $u = x - y.$ equation satisfied by $u$ is simpler than the one you have. $\endgroup$ – abel Nov 24 '14 at 2:50
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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 :

enter image description here

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

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