$xy'-y=0$ 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

Without log function
$$y'x-y=0$$
for $x \ne 0$
$$\frac {(y'x-y)}{x^2}=0$$
$$\left(\frac yx \right)'=0$$
Integrate
$$\frac yx= K$$
$$y(x)=Kx$$
