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I Have differential equation with module which looks:

$$ |y| \sqrt{1-u}dy - y^{2} du = 0 $$

where $ u = \frac{x}{y}, u \leq 1$

Next I'm trying divide equation both side by: $ y^{2} \cdot \sqrt{1-u} $ and I'm obtaning result:

$$ \pm \frac{dy}{y} - \frac{du}{\sqrt{1-u}} = 0 $$

where of course $ \pm $ is Due to the absoulte value.

After calculating two integrals:

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

$$ \int \frac{du}{\sqrt{1-u}} = -2(1-u)^{\frac{1}{2}} +C $$

I have result like:

$$ \pm ln|y| + 2(1-\frac{x}{y})^{\frac{1}{2}} = 0$$

In case where $y > 0$ ( $y = 0$ because of dividing by $y^2$ ) my score is:

$$ y = d e^{-2(1-\frac{x}{y})^{\frac{1}{2}}} $$ where: $ d = e^{C} $ and $ x < y$

I second case I have a result:

$$ y = d e^{-2(1-\frac{x}{y})^{\frac{1}{2}}} $$ where: $ d = - e^{-C}$, $x < y$

I don't know why in book from exercice from is in that case result like:

$$ y = c e^{2(1-\frac{x}{y})^{\frac{1}{2}}} $$ where: $ y < 0 $, $ x > y$

Why here is $x > y$ instead $ x< y $ like in first case?

In my opinion should be:

$$ \frac{x}{y} < 1 $$ $$ x < y $$

Could someone explain me why? Is correct my answer in second case? I will be greatful for the answers. Best regards

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    $\begingroup$ How can the book say $x>y$ when at the start it says $\frac xy=u\leq1$? Seems like there's something wrong there $\endgroup$ – John Doe May 6 '17 at 18:06
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    $\begingroup$ @JohnDoe $$-1>-2\qquad\frac{-1}{-2}\le1$$ $\endgroup$ – Did May 7 '17 at 4:39

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