Differential Equation with Substitution I have this Differential Equation:
$$(2x^2 + y^2) \frac{dy}{dx} = 2xy$$
and I solved it by substituting $y=vx$ and $\frac{dy}{dx}=\frac{dv}{dx} (x +v)$
I found a result of $y^2(\ln(y) +C)=x^2$, but since it must be an implicit solution for $y$ I don't know whether I should leave it in this form. Also I hope I didn't do anything wrong in the proccess. Any help would be much appreciated
 A: There is a typing mistake in :
$y=vx$ and $\frac{dy}{dx}\neq\frac{dv}{dx} (x +v)$
$$\frac{dy}{dx}=v+x\frac{dv}{dx}$$
With this, you got to :
$$x^2=y^2(c+ln(y))$$
That's OK. You can let it on the implicit form (or an equivalent form).
One can express $y(x)$ thanks to a special function called Lambert W function.
$$y=\pm \frac{\sqrt 2 \: x}{\sqrt{W(C x^2)}}$$
A: here is another way to solve $$\frac{dy}{dx} = \frac{2xy}{2x^2 + y^2}.$$  split the differential equation into two as $$\frac{dy}{dt} = \frac y{2x^2+y^2}, \frac{dt}{dx}=2x.$$
solving for $x,$ we get $$x^2 = t +C. \tag1$$ and the differential equation for $y$ is $$ \frac{dt}{dy} = \frac{2x^2+y^2}{y} = \frac{2(t + C)}{y}  + y.$$ which can be solved as $$ t + C = y^2\ln y +By^2.\tag 2$$ eliminating $t$ between $(1)$ and $(2)$ gives you $$x^2 = y^2 \ln y+By^2, \text{ where  $B$ is an arbitrary constants.  } $$
i can't figure out where i am losing an $\ln y.$ 
edit: thanks to narasimham. i found the error at the very beginning. i copied the wrong equation. it is good now.
