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I was trying to solve this differential equation but can't figure out the final integral I get by variable separable method

The equation is $$ x^3 \, y' = y^3 + y^2 \, \sqrt{y^2-x^2} $$

I got the integral $$ \frac{dv}{v^3 + v^2 \sqrt{(v^2 - 1)} - v} $$ but can't figure out how to solve it.

The Euler substitution $v= \frac{u^2 + 1}{2u}$ could work but I can't seem to proceed with it.

Any help with this problem is much appreciated.

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  • $\begingroup$ Did you try using Wolfram Alpha ? $\endgroup$ – Claude Leibovici May 11 '14 at 9:26
  • $\begingroup$ Yes I tried Wolfram alpha . This seems to be an easy integral but I can't seem to figure it out. The final integral that I've posted needs to be solved but how ? $\endgroup$ – Curious May 11 '14 at 13:38
  • $\begingroup$ I think that Jeb made it ! $\endgroup$ – Claude Leibovici May 11 '14 at 14:05
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I think I see a really fast answer to this. Divide numerator and denominator by $v^3$.

$$\int\frac{v^{-3}dv}{1-v^{-2}+\sqrt{1-v^{-2}}}=\int\frac{v^{-3}dv}{(\sqrt{1-v^{-2}}+1)\sqrt{1-v^{-2}}}$$ $$t=\sqrt{1-v^{-2}}+1,dt=\frac{2v^{-3}dv}{2\sqrt{1-v^{-2}}}=\frac{v^{-3}dv}{\sqrt{1-v^{-2}}}$$

This drastically reduces the integral to $\int\frac{dt}t$.

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  • $\begingroup$ you got the integrand wrong \frac{dv}{v^3 + v^2 \sqrt{(v^2 - 1)} - v} $\endgroup$ – Curious May 12 '14 at 10:36
  • $\begingroup$ @Curious Oops. Don't know where those pluses came from. I'll fix it. Fortunately, it's just cosmetic. Actually, it wouldn't come out that cleanly if it were a plus under the radical. $\endgroup$ – Mike May 12 '14 at 17:22
  • $\begingroup$ @Curious It should be correct now if you want to give it another look. $\endgroup$ – Mike May 12 '14 at 17:27
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    $\begingroup$ @Curious Or you do as I did. $\frac{v^2\sqrt{v^2-1}}{v^3}=\frac{v^2}{v^2}\times\sqrt{\frac{v^2-1}{v^2}}= \sqrt{1-v^{-2}}$ $\endgroup$ – Mike May 14 '14 at 14:22
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    $\begingroup$ @Curious Added a small step, factoring the denominator. Does this help? $\endgroup$ – Mike May 15 '14 at 17:08
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Simply use a partial fraction decomposition to obtain:

$$ \int \frac{ dx}{x^3 + x^2 \sqrt{ x^2 -1} - x } = \int \frac{ \sqrt{x^2 -1} }{2 (x-1) } - \frac{ \sqrt{ x^2 - 1}}{ 2 (x+1)} - \frac{1}{x} dx$$

Use $u$ substitute with $x-1$ and $x+1$ since $x^2-1 = (x+1)(x-1)$. I think that should do it.

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  • $\begingroup$ Well done ! Thanks. $\endgroup$ – Claude Leibovici May 11 '14 at 14:04
  • $\begingroup$ @Jeb The partial decomposition doesn't yield the integral mentioned , right ? I tried doing it but it is (x-sqrt{x^2-1})/{x*sqrt(x^-1)} $\endgroup$ – Curious May 11 '14 at 14:34
  • $\begingroup$ That isn't right, my suggestion puts it in the form $ \sqrt{ (1 \pm 2/u )} du$ which looks like something I've done in the past. I don't remember how to solve it explicitly though $\endgroup$ – Jeb May 11 '14 at 14:40
  • $\begingroup$ The partial decomposition mentioned does not yield L.H.S explicitly. Correct me if I'm wrong $\endgroup$ – Curious May 11 '14 at 14:43
  • $\begingroup$ It does, wolframalpha.com/input/… $\endgroup$ – Jeb May 11 '14 at 14:46
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$$ x^3 \, y' = y^3 + y \, \sqrt{y^2-x^2} $$ Let (y/x)=t, then: y = v*x; y' = x*v' + v

use this substitution

...and v^3 + (v^2)sqrt[v^2 - 1] - v = v[(v^2 - 1) + sqrt(v^2 - 1)]

v^2 - 1 = p^2 => vdv = pdp

and you will get:

(dp)/[(p^2 +1)*(p + 1)] = dx/x

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  • $\begingroup$ You have put (y/x)*sqrt[(y/x)^2-1] but it was (R.H.S)/x^3. So it should be (y/x^2)*sqrt[(y/x)^2-1] + (y/x)^3 = y' $\endgroup$ – Curious May 11 '14 at 13:30
  • $\begingroup$ I made ​​a start on the integral that you have specified. there was v^3 + (v^2)*sqrt[(v^2 - 1) - v] then my solving, of course, is not suitable for the initial condition $\endgroup$ – Alenka May 11 '14 at 14:09
  • $\begingroup$ you've got y'= (y/x)^3 + (y/(x^2))*sqrt((y/x)^2 -1) how u got v^3 + (v^2)*sqrt...? $\endgroup$ – Alenka May 11 '14 at 14:36
  • $\begingroup$ v^3 + (v^2)*sqrt[(v^2 - 1)]- v $\endgroup$ – Curious May 11 '14 at 14:47
  • $\begingroup$ y/(x^2) is not equal to v^2 $\endgroup$ – Alenka May 11 '14 at 15:11

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