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The following differential equation is given: $$\frac{dy}{dx}=\frac{y^2}{x}$$

Separating variables and integrating: $$\int y^{-2} dy = \int x^{-1} dx$$ $$-y^{-1}=\ln|x|+c$$ $$y=\frac{1}{-c-\ln |x|}$$

But the solution is given as: $$y=\frac{1}{-c-\ln x}$$

How can this omission of the modulus be explained?

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    $\begingroup$ The absolute value is sometimes skipped for convenience, if the domain is assumed to be $(0,\infty)$. $\endgroup$ – Dylan Jul 19 '18 at 16:22
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The solution given is for $x>0$ only. There is also the solution $$y=\frac{1}{-c-\ln(-x)}$$ for $x<0$ only. No solution may cross $x=0$.

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    $\begingroup$ In particular, thanks to the discontinuity, the "c" in each half can be specified independently. $\endgroup$ – IanF1 Jul 19 '18 at 17:31

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