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I'm trying to solve a differential equation but I get to the point in which I have to solve this:
$$\int {\frac {y(x)' (1-y(x)^2)}{y(x)^3}} dx$$
I don't understand how to do it. I think it's related to some property I don't know, but the $y(x)'$ is throwing me off. I tried looking at Wolfram's step by step but I don't understand it.
Thanks for any help.

Context: Original function to solve is $(v^3) du + (u^3-uv^2) dv = 0$.
I used a variable change: $y = \dfrac uv$. I'm not sure if it's correct or anything though.

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    $\begingroup$ What is the differential equation you're trying to solve? How does this integral come up? If you're using separation of variable there shouldn't be a $y'$ in any integral. $\endgroup$ – DanZimm Jun 6 '14 at 6:26
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    $\begingroup$ Hint: use the chain rule. $\endgroup$ – David H Jun 6 '14 at 6:27
  • $\begingroup$ Also welcome to MSE! $\endgroup$ – DanZimm Jun 6 '14 at 6:27
  • $\begingroup$ @DavidH Oh, I always forget the chain rule for integrals. Thanks! $\endgroup$ – Paula Jun 6 '14 at 6:32
  • $\begingroup$ And @DanZimm I used first a change of variables, edited the question for context if you're curious :). And thanks for the welcome! $\endgroup$ – Paula Jun 6 '14 at 6:33
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Using the obvious change of variables $y=y(x) \implies dy=y'(x)\,dx$, your integral becomes:

$$\int {\frac {y(x)' (1-y(x)^2)}{y(x)^3}} dx=\int {\frac {1-y^2}{y^3}} dy\\ =\int y^{-3}dy-\int y^{-1}dy\\ =-\frac12 y^{-2}-\log{y}+constant$$

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  • $\begingroup$ Oh, that was very clear, thanks! $\endgroup$ – Paula Jun 6 '14 at 7:42
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The differential equation simplifies a lot if you use $$u(v)=\frac{v}{y(v)}$$ (which is basically the reciprocal of what you chose).

After replacement and simplifications (there are several), you should arrive to $$1-v y(v) \frac{dy(v)}{dv}=0$$ for which the solution is simply $$y(v)= \sqrt{c_1+2\log (v)}$$ from which $$u(v)=\frac{v}{\sqrt{c_1+2 \log (v)}}$$

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  • $\begingroup$ Thanks for the answer! It's easier that way. $\endgroup$ – Paula Jun 6 '14 at 7:44
  • $\begingroup$ You are really welcome. I don't have any merit since I noticed that after your change of variable the equation was almost as complex as the initial one. So, I took the reciprocal ! Cheers and again welcome to this site. $\endgroup$ – Claude Leibovici Jun 6 '14 at 7:56

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