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$(y^2\sqrt{x-x^2y^2}-y)dx - 2xdy=0$

Change it into this $$y'=\frac{y^2\sqrt{x-x^2y^2}-y}{2x}$$

Square root disables a lot of methods.

It isn't a total differential, I've tried. Quasi-homogeneous $y=z^a$ is only thing I haven't ruled out but I don't know how to determine $a$ because of the square root. If it is quasi-homogeneous how do I find $a$?

Desperado
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2 Answers2

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All first order equation of the first $M(x,y)dx+N(x,y)dy=0$ can be integrated, provided that you can find the right integrating factor. For this equation $(y^2\sqrt{x-x^2y^2}-y)dx - 2xdy=0$, the integrating factor is $$ \frac{1}{xy^2\sqrt{x-x^2y^2}}.$$ Multiplying both sides with this integrating factor, we get $$ \left(\frac{1}{x}-\frac{1}{xy\sqrt{x-x^2y^2}}\right)dx-\frac{2}{y^2\sqrt{x-x^2y^2}}dy=0.$$ This is now a total differential and integrating on both sides leads to $$ \ln x-\frac{2}{y}\sqrt{\frac{1-xy^2}{x}}=C.$$ The solution then can be obtained by solving $y$ from the algebraic equation.

yhhuang
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  • I haven't done any problems with an integrating factor that's a function x and y. Is $u(x,y)= e^{\int{G(x) dx}}e^{\int{F(y) dy}}.$ , where U is the integrating factor, how you got the solution? The method I found – Desperado Dec 21 '15 at 10:34
  • In general there is no systematic way to find the integrating factor of the equation $M(x,y)dx+N(x,y)dy=0$. I tried the hard way, by looking different combination of the terms. You have to solve a partial differential equation in the method you mentioned, which could be more difficult. – yhhuang Dec 21 '15 at 11:15
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This is separable. The hint is in the terms outside the square root. But first, lets talk a bit about where this is defined.

Clearly we must have $x \ge x^2y^2$. Which also means we must have $x\ge 0$. Furthermore, you can check that $x=0$ and $y=0$ are both solutions. So we will be looking for solutions in the $x>0$ region.

Lets write the differential as $$ y^2\sqrt{x(1-xy^2)}\ dx = ydx + 2xdy. $$ Hopefully, you can see the the left hand side suggests that we can make the substitution $z = xy^2$ (check it), with $z\in (0,1)$. What comes out the other side is $$ \frac{dx}{x} = \frac{dz}{\sqrt{z^3(1-z)}}. $$ We can solve this to get $$ \ln x = -2\sqrt{\frac{1}{z} -1} +c. $$ for some constant $c$.

Working our way back up we should find $$ y = \pm \frac{2}{\sqrt{x}\sqrt{(c - \ln x)^2 +4)}}. $$

MrSlunk
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