Absolute Value in Linear Differential Equation

I have a question about dropping the absolute value sign when solving a linear differential equation.

If $y'-y/x=1$

Integrating Factor $=e^{\int{-1/xdx}}=e^{-lnx}=1/x$

$y/x-y/x^2=1/x$

$[y/x]'=1/x$

$y/x=\int{1/x dx}+C$

$y=xln\lvert{x}\rvert+Cx$

However textbook solutions and Mathematica show:

$y=xlnx+Cx$

I read somewhere that because you multiple both sides by the integrating factor you can drop the absolute value. My question is when you perform the final integration, how is it that you are able to drop the absolute value again? In general, what are the guidelines for dropping the absolute value sign?

• If there is no particular specification, then $y=x\ln(x)+Cx$ is only a part of the solution and the complete solution is $y=x\ln\lvert{x}\rvert+Cx$. But if a specified condition implies $x>0$ then $y=x\ln(x)+Cx$ is the solution. So, knowing the initial conditions is important to give a definitive answer. – JJacquelin Aug 14 '14 at 6:48