We want to solve the differential equation $y'=\frac{1-x+y}{x-y}$, What i did is define $z=x-y$, and then $y'=1-z'$, so overall we have
$1-z'=\frac{1-z}{z}$, or in other words $z'=\frac{2z-1}{z}$ This can be solved because $z$ and $x$ are separable:
$\int \frac{z}{2z-1}dz=\int dx$
I don't know any method to integrate the left side, so I'll be grateful if someone can show an analytic way to do that, but i managed to guess the integral: $\frac{1}{2}z+\frac{1}{4}\ln (2z-1)=x+c$
Now to revert back to our original $y$:
$\frac{1}{2}(x-y)+\frac{1}{4}\ln (2(x-y)-1)=x+c$
Is this the best we can do? Is there a way to find an explicit form for $y$? or an implicit equation like this is the best solution we can hope for in this situation
There is a simple way to integrate the integral you have:
$$\frac{z}{2z-1} = \frac{1}{2}\frac{2z}{2z-1} = \frac12\frac{(2z-1)+1}{2z-1} = \frac12\left(1 + \frac1{2z-1}\right).$$
As for solving your equation for $y$, no, I do not think it is possible to get a closed form of $y$. I am not sure (there are theorems that tell you, but I don't know them), but the rule of thumb is that if an equation contains the variable both in the exponent and in line, or in your case, in the logarithm and in line, then a closed form solution is not likely.