Is okay to have different solution to differential equation? Suppose I have the following differential equation: 
$ydx - xdy - dx = 0$
Now, I could divide it by Integrating factor $x^2$ to get:
$(xdy - ydx)/(x^2) - dx/x^2 = 0$
Use the inspection rule to get:
$d(y/x) + dx/x^2 = 0$
Then Integrate this exact differential equation to get the answer:
$(y - 1)/x = c$
That's seems okay..
But what if I had divided it by $y^2$?
The solution I then get would be:
$d(x/y) + dx/y^2 = 0$
Is that completely fine to have different Integrating factors or should you always pick $x^2$?
 A: I believe you found the integration factor $\frac{1}{x^2}$ using the typical way, although I didn't check. This gives a correct solution. 
The $\frac{1}{y^2}$ won't work since the second one is not an exact equation. For it to be exact like the first one, you need
$$d(\frac{x}{y})-\frac{dy}{y^2}=0$$
whereas you have 
$$d(\frac{x}{y})-\frac{dx}{y^2}=0$$
which is not exact since the $\frac{dx}{y^2}$ term is not a differential. 
And by the way, since this is a linear differential equation, it should have one solution. An alternative way to solve it is to use separation of variable:
$$y-1-x\frac{dy}{dx}=0\\
\frac{dx}{x}=\frac{dy}{y-1}\\
\ln |x|=\ln |y-1|+C\\
x=c (y-1)$$
Edit:
There is a different integration factor though, which is $\frac{1}{(y-1)^2}$. The integration factors are obtained by either 
$$\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N}\tag1$$
or 
$$\frac{\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}}{M}\tag2$$
They should give you the same result.
Edit 2:
First of all, a differential equation is exact if
$$\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=0$$
If not, we can multiply by an integration factor if (1) only depends on $x$. In that case, the integration factor is $$e^{\int (1) dx} $$
where (1) means the expression in (1). Otherwise we check whether (2) depends only on $y$. If that is true, the integration factor is $$e^{\int (2) dx}$$ These steps give you an exact equation. 
