# Differential equation Euler substitution

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.

• Did you try using Wolfram Alpha ? – Claude Leibovici May 11 '14 at 9:26
• 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 ? – Curious May 11 '14 at 13:38
• I think that Jeb made it ! – Claude Leibovici May 11 '14 at 14:05

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 $\int\frac{dt}t$.

• you got the integrand wrong \frac{dv}{v^3 + v^2 \sqrt{(v^2 - 1)} - v} – Curious May 12 '14 at 10:36
• @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. – Mike May 12 '14 at 17:22
• @Curious It should be correct now if you want to give it another look. – Mike May 12 '14 at 17:27
• @Curious Or you do as I did. $\frac{v^2\sqrt{v^2-1}}{v^3}=\frac{v^2}{v^2}\times\sqrt{\frac{v^2-1}{v^2}}= \sqrt{1-v^{-2}}$ – Mike May 14 '14 at 14:22
• @Curious Added a small step, factoring the denominator. Does this help? – Mike May 15 '14 at 17:08

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.

• Well done ! Thanks. – Claude Leibovici May 11 '14 at 14:04
• @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)} – Curious May 11 '14 at 14:34
• 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 – Jeb May 11 '14 at 14:40
• The partial decomposition mentioned does not yield L.H.S explicitly. Correct me if I'm wrong – Curious May 11 '14 at 14:43
• It does, wolframalpha.com/input/… – Jeb May 11 '14 at 14:46

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

• 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' – Curious May 11 '14 at 13:30
• 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 – Alenka May 11 '14 at 14:09
• you've got y'= (y/x)^3 + (y/(x^2))*sqrt((y/x)^2 -1) how u got v^3 + (v^2)*sqrt...? – Alenka May 11 '14 at 14:36
• v^3 + (v^2)*sqrt[(v^2 - 1)]- v – Curious May 11 '14 at 14:47
• y/(x^2) is not equal to v^2 – Alenka May 11 '14 at 15:11