$xy'-y=0$ when $x<0$ 
How to solve $xy'-y=0$ when $x<0$?

It seem to be a simple equation, but it is confusing when $x<0$ 
$$\frac{y'}{y}=\frac{1}{x}$$
Now to integrate the both sides must I assume that $y$ is positive then 
$$\ln y =- \ln(-x)+c$$so 
$$y(x)=e^{-\ln(-x)}$$ 
Am I right? what if the assumption on $y$ is false? 
This equation has a solution over all $\mathbb{R}$?
Thanks in advence

You do not presume $y$ to be positive or negative. Rather,  $$\frac{y'}{y}=\frac{1}{x} \implies \ln(\color{red}{|y|}) = \ln(-x) + c \implies |y| = e^{\ln(-x)+c} \implies  y(x) = C x,$$
where $C$ can be positive or negative.
Also, the above result holds for all values of $x$, positive or negative. To verify, observe that $$y' = C \implies x y' - y = 0.$$
