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I have a differential equation:

$$ \frac{dy}{y} + \frac{1+2u^{2}}{2u}\, du = 0 $$ where $u = \frac{x}{y} $

Because:

$$ \int \frac{1+2u^2}{2u} du = \frac{1}{2} \ln|u| + \frac{u^2}{2} + C $$

and:

$$ \int \frac{dy}{y} = \ln|y| + C $$

So, I'm obtaining equation:

$$ \ln|y| + \frac{1}{2} \ln |u| + \frac{u^{2}}{2} = C $$

Next: $$ \ln|y| + \frac{1}{2} \ln |\frac{x}{y}| + \frac{x^2}{2y^2} = C $$

Next, I'm dividing equation both sides by $2$ and I'm obtaining:

$$ \ln|y^2| + \ln|\frac{x}{y}| + \frac{x^2}{2y^2} = 2 C$$

$$ \ln|x y| + \frac{x^2}{y^2} = d$$ where $d = 2C$ (other constant)

I don't know why in book from which this equation froms answer is without module:

$$ \frac{x^2}{y^2} + \log(xy) = c, x \neq 0, y \neq 0 $$

Could someone exaplain why book's answer dosen't have module(absoulte value)? I would be greatful for helps Best reagards ;)

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  • $\begingroup$ Yes, it means the same. $\endgroup$ – Krzysztof Michalski May 13 '17 at 11:59
  • $\begingroup$ Then I'm obtaining following result: $$ de^{ \frac{-y^{2}}{x^{2}} } = \frac{y^{3}}{x} $$, where $d = +- e^{2C}$ but it's the same like answer from book $\endgroup$ – Krzysztof Michalski May 13 '17 at 12:10
  • $\begingroup$ Orginally it is differential equation: $$ (2x^{2} y + y^{3})dx + (xy^{2}-2x^{3})dy = 0 $$ wich I'm trying to solve by substitution: $x = uy, dx = udy + ydu$ $\endgroup$ – Krzysztof Michalski May 13 '17 at 12:20
  • $\begingroup$ The answer without the absolute value looks like a typo. If a solution starts in the first quadrant ($x > 0$ and $y > 0$) it will stay there (so no absolute value is needed), but in general the $\ln(xy)$ term is non-real, while the $\ln |xy|$ term in your formula makes sense off the coordinate axes, and generalizes your book's formula. $\endgroup$ – Andrew D. Hwang May 13 '17 at 12:27

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