# differential equation with module

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

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