# differential equation. Problem with module

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 ;)

• Yes, it means the same. – Krzysztof Michalski May 13 '17 at 11:59
• 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 – Krzysztof Michalski May 13 '17 at 12:10
• 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: $x = uy, dx = udy + ydu$ – Krzysztof Michalski May 13 '17 at 12:20
• 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. – Andrew D. Hwang May 13 '17 at 12:27