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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?

Thanks in advance

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    $\begingroup$ Just out of curiosity, was this the full problem or was there an initial condition? $\endgroup$ – Mike Aug 14 '14 at 6:12
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    $\begingroup$ 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. $\endgroup$ – JJacquelin Aug 14 '14 at 6:48
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    $\begingroup$ Wow this is embarrassing apparently I misread the solution but you guys are right since there's no initial condition, it does have the absolute value. I find it interesting that Mathematica drops the absolute value with no explanation. Thanks and sorry for kind of wasting your time. $\endgroup$ – bkmoney Aug 14 '14 at 7:40

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