Some double angle identity to solve $(2x^{2}+y^{2})\frac{dy}{dx}=2xy$? For some reason, I cannot see a clever way to solve this (I know the way of doing it like in Wolframalapha) but I am pretty sure there is a double angle identity to crack this puzzle. Could someone hint a bit to get this puzzle onwards?
Firstly, I thought to use some rules such as $(x+y)^{2}=x^{2}+2xy+y^{2}$ or $(x-y)^{2}=x^{2}-2xy+y^{2}$ but I think some trigonometric substitution could solve this problem.
 A: If we write the equation as,
$$\frac{{dy}}{{dx}} = \frac{{2xy}}{{2{x^2} + {y^2}}}$$
and then divide through $x^2$ we will get:
$$\frac{{dy}}{{dx}} = \frac{{2\dfrac{y}{x}}}{{2 + {{\left( {\dfrac{y}{x}} \right)}^2}}}$$
This suggests that we simplify the previous equation in terms of 
$$f\left( v \right) = \frac{{2v}}{{2 + {v^2}}}$$
So putting 
$$\eqalign{
  & \frac{y}{x} = v  \cr 
  & y = vx  \cr 
  & y' = v'x + v \cr} $$
We get
$$\frac{{dv}}{{dx}}x + v = \frac{{2v}}{{2 + {v^2}}}$$
Then
$$\eqalign{
  & \frac{{dv}}{{dx}}x =  - \frac{{{v^3}}}{{2 + {v^2}}}  \cr 
  & \frac{{dx}}{x} =  - \frac{{2 + {v^2}}}{{{v^3}}}dv  \cr 
  & \frac{{dx}}{x} = \left( { - \frac{2}{{{v^3}}} - \frac{1}{v}} \right)dv \cr} $$
Upon integration we have:
$$\log x + C = \frac{1}{{{v^2}}} - \log v$$
Let's substitute back
$$\eqalign{
  & \log x + C = \frac{{{x^2}}}{{{y^2}}} - \log \frac{y}{x}  \cr 
  & \log x + C = \frac{{{x^2}}}{{{y^2}}} - \log y + \log x  \cr 
  & \log y = \frac{{{x^2}}}{{{y^2}}} - C  \cr 
  & y = {C_1}\exp \left( {{x^2}{y^{ - 2}}} \right) \cr} $$
You can find $y$ in terms of $x$, but I don't think the inverse is possible, at least with everyday functions.
$$y\sqrt {\log y + C}  = x$$
Ok, using the Lambert W we have
$${y^2}\left( {\log y + C} \right) = {x^2}$$
Use the exponential:
$${e^{{y^2}}}y{e^C} = {e^{{x^2}}}$$
Square and multiply by two
$$2{y^2}{e^{2{y^2}}}{e^{2C}} = 2{e^{2{x^2}}}$$
Use the Lambert W
$$2{y^2} = W\left( {\frac{{2{e^{2{x^2}}}}}{{{e^{2C}}}}} \right)$$
$$y = \sqrt {\frac{1}{2}W\left( {\frac{{2{e^{2{x^2}}}}}{{{e^{2C}}}}} \right)} $$
Another aproach would be 
$$\eqalign{
  & \log y + C = \frac{{{x^2}}}{{{y^2}}}  \cr 
  & y{e^C} = {e^{\frac{{{x^2}}}{{{y^2}}}}}  \cr 
  & {y^2}{e^{2C}} = {e^{2\frac{{{x^2}}}{{{y^2}}}}}  \cr 
  & 2\frac{{{x^2}}}{{{y^2}}}{y^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}}  \cr 
  & 2{x^2}{e^{2C}} = 2\frac{{{x^2}}}{{{y^2}}}{e^{2\frac{{{x^2}}}{{{y^2}}}}}  \cr 
  & W\left( {2{x^2}{e^{2C}}} \right) = 2\frac{{{x^2}}}{{{y^2}}}  \cr 
  & {y^2} = \frac{{2{x^2}}}{{W\left( {2{x^2}{e^{2C}}} \right)}}  \cr 
  & y = \frac{{\sqrt 2 x}}{{\sqrt{W\left( {2{x^2}{e^{2C}}} \right)}}} \cr} $$
A: Hint: it's an homogeneous differential equation.
A: Rewrite equation into form :
$$\frac{dy}{dx}=\frac{2xy}{2x^2+y^2}$$
Substitute :
$$z =\frac{y}{x} \Rightarrow y'=xz'+z$$
Therefore :
$$xz'+z=\frac{2z}{2+z^2} \Rightarrow xz'=\frac{-z^3}{2+z^2} \Rightarrow \int \frac {2+z^2}{z^3} \,dz= -\int \frac {dx}{x} $$
