Find value of $xy\sqrt{y^2 - x^2}$ for the given differential equation. 
If $(y^3 - 2x^2y)dx + (2xy^2 - x^3)dy = 0 $ , then prove that the value of $xy\sqrt{y^2 - x^2}$ is a constant. 

This is what I've tried : 
$$ y(y^2 - 2x^2) dx + x(2y^2 - x^2) dy = 0 \\
\cfrac{dy}{dx} = \cfrac{(2x^2 - y^2)y}{(2y^2 - x^2)x} \\
$$
I then tried substituting $y/x = v$ 
Further, solving it, I got : 
$$xy\sqrt{y^2 - x^2} = \cfrac{x^{4/3}}{y^{4/3}}$$
But, I'm not sure how to come with a constant at RHS. 
 A: When you substitute the change of variables you obtain:
$$ xu' = u \left( \frac{2-u^2}{2u^2-1} -1 \right), $$
which upon integration and taking exponential in both sides gives (I used Mathematica):
$$ x e^C = \frac{1}{u^{1/3} ({1-u^2})^{1/6}}, $$ where $e^C :=K$ is a constant of integration. Substitute back $u = y/x$ to find:
$$ K x =  \frac{1}{\left(\frac{y}{x} \right)^{1/3} \left({1-\frac{y^2}{x^2}} \right)^{1/6}},$$ now cube everything to come up with:
$$ K x^2 y \left( 1-y^2/x^2  \right)^{1/2} = K x y \sqrt{x^2-y^2 }=1, $$ which is equivalent to the desired result.
Hope this helps!
A: From the very beginning, use $y=x\,v$, $y'=x v'+v$ and replace in the original equation. You will get a common factor $x^3$; simplify to get $$x \left(2 v^2-1\right) v'+3 v \left(v^2-1\right)=0$$ which is separable becoming $$\frac 1x \frac{dx}{dv}=-\frac{2 v^2-1}{3 v \left(v^2-1\right)}$$ Use partial fraction decomposition to get $$\frac 1x \frac{dx}{dv}=-\frac{1}{3 v}-\frac{1}{6 (v+1)}-\frac{1}{6 (v-1)}$$ Integrate both sides $$\log(x)+C=-\frac{1}{6} \log \left(1-v^2\right)-\frac{1}{3}\log (v)$$ that is to say $$Cx^6=\frac{1}{v^2(1-v^2)}$$ Make the cross product, replace $v$ by $\frac y x$ to obtain $$C x^2 y^2 (x^2-y^2)=1$$ from which the result arrives.
