I have this differential equation in an exercise:

$$y' = \frac{y}{\sqrt{1-x^2}}$$

I solved it like this:

$$\frac{dy}{dx} = \frac{y}{\sqrt{1-x^2}}$$

$$\frac{1}{y}dy = \frac{1}{\sqrt{1-x^2}}dx$$

$$\int \frac{1}{y} \,dy = \int \frac{1}{\sqrt{1-x^2}} \,dx$$

$$\ln \,\lvert\,y\,\lvert + C_1 = \arcsin\,x + C_2$$

$$\lvert\,y\,\lvert\,= {e}^{\arcsin\,x + C_3}$$



However, the solution given in my textbook is


How to get rid of the absolute value? Did I do something wrong? Some of the other exercises indicates when $y > 0$ but this one has no such indication. I guess it can be deducted?

  • 3
    $\begingroup$ $C$ is an arbitrary constant. $\endgroup$ – Donald Splutterwit Sep 24 '17 at 19:52
  • $\begingroup$ R.H.S. is always positive $\endgroup$ – haqnatural Sep 24 '17 at 19:53
  • $\begingroup$ always $e^x>0$ . $\endgroup$ – Nosrati Sep 24 '17 at 19:59
  • $\begingroup$ When you divided by $y$ you also missed the solution $y \equiv 0$ which corresponds to $C=0$. $\endgroup$ – Daniel Schepler Sep 24 '17 at 20:01

Since $C$ is an arbitrary constant it can have a positive or a negative value, (e.g $C=\pm 2$).


I understood thanks to Emilio Novati's answer but the confusion comes from the difference between my constant and the one of the given answer. The $C$ coming from my steps is positive (since ${e}^{C_3}$ can't be negative) but I can simply use another letter such as $$K = ±C$$

So I get



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