# How to approach this simple differential equation?

$(1-y)dx - (1-x)dy = 0$

I know that for it to be exact the sign should be positive, and I've also seen one example (with -) where the approach was to make the sign positive to make it exact. However, I think it makes more sense to approach it as a separable equation:

$(1-y)dx = (1-x)dy$

$\frac{1}{1-x} = (\frac{1}{1-y}) \frac{dy}{dx}$

This should be pretty easy, but I'm checking it with some software and it shows this instead:

$\frac{1}{x-1} = (\frac{1}{y-1}) \frac{dy}{dx}$

The rest of the steps are only different because of the signs, but I'm confused because I don't understand how $(1-x)$ and $(1-y)$ became $(x-1)$ and $(y-1)$ when the software solved it. This looks like it could be a very basic thing that I never learned.

1.-Am I right to approach this as a separable equation?

2.-How does that sign change happen?

2 - The software is, for admittedly arbitrary reasons, multiplying both sides of the equation by $-1$. This has no real effect.
$$(1-y)dx-(1-x)dy=0\implies \frac{dy}{1-y}=\frac{dx}{1-x}\implies \log|1-y|=\log|1-x|+C\text{(onstant)}$$