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I know the basics about separable, exact, and homogeneous, but several things about this one confuse me. The equation is:

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

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?

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1 - Separability is exactly the right way to approach this problem. You might also find it informative to draw a slope field.

2 - The software is, for admittedly arbitrary reasons, multiplying both sides of the equation by $-1$. This has no real effect.

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Hint:

$$(1-y)dx-(1-x)dy=0\implies \frac{dy}{1-y}=\frac{dx}{1-x}\implies \log|1-y|=\log|1-x|+C\text{(onstant)}$$

and mind the definition domains and stuff. I don't know what you meant with "exact has to be positive" ...

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