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I have to find the solution to the differential equation

$$(y^2-1)+2(x-y(1+y)^2)y'=0$$

So far I've only learned how to solve equations of the form $$y'+p(x)y=q(x)$$ And second order equations with constant coefficients. This equation that I have to solve right now does not seem to be of any of these $2$ forms so I don't understand how I can solve this. I know I'm supposed to show some attempted solutions I tried but frankly I just have no idea how to even start here.

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  • $\begingroup$ why have you written $(y^2-1)$ in brackets? $\endgroup$ – Dr. Sonnhard Graubner May 9 '15 at 12:04
  • $\begingroup$ @Dr.SonnhardGraubner I don't know, it was that way in my book. I think they wrote it like that the suggest something, but I don't know what they were going for... $\endgroup$ – user2520938 May 9 '15 at 12:05
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Hint : instead of searching for $y(x)$, search for the inverse function $x(y)$ : $$(y^2-1)\frac{dx}{dy}+2(x-y(1+y)^2)=0$$ $$(y^2-1)\frac{dx}{dy}+2\:x=2y(1+y)^2$$ It is a linear ODE that can be solved for $x$.

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$\bf hint:$ write it as an exact differential equation $$\frac{y-1}{y+1}\frac{dy}{dx} + \frac{2}{(y+1)^2} = 2y.$$


here are some of the steps i left for you to do:

(a) write the differential equation as $$\frac{dx}{dy} = \frac{2\left(y(y+1)^2-x\right)}{y^2 - 1} $$

(b) $$\frac{dx}{dy} + \frac{2x}{y^2 - 1} = \frac{2y(y+1)^2}{y^2 - 1} $$

(c) show that $$\frac{y-1}{y+1}$$ is an integrating factor for the differential equation in (b).

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  • $\begingroup$ I have absolutely no idea what you're doing here. I'm interested in how to solve it, not in actually solving it. So just giving me some random hints without explaining what you're doing really doesn't help me $\endgroup$ – user2520938 May 9 '15 at 12:18
  • $\begingroup$ @user2520938, please see my edit. $\endgroup$ – abel May 9 '15 at 12:24
  • $\begingroup$ Thanks. It's just that I'm really new to this, so although I normally like getting hints more than getting full answers, in this case I really needed it getting spelled out for me:p. $\endgroup$ – user2520938 May 9 '15 at 13:37
  • $\begingroup$ @user2520938, no problem. i assumed you knew how to find the integrating factor. $\endgroup$ – abel May 9 '15 at 13:40

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