# Some double angle identity to solve (2x^{2}+y^{2}) \frac{dy}{dx}=2xy?

Some double angle identity to solve $\left(2{x}^{2}+{y}^{2}\right)\frac{dy}{dx}=2xy?$
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Mary Herrera
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:
$\frac{dy}{dx}=\frac{2\frac{y}{x}}{2+{\left(\frac{y}{x}\right)}^{2}}$
This suggests that we simplify the previous equation in terms of
$f\left(v\right)=\frac{2v}{2+{v}^{2}}$
So putting
$\frac{y}{x}=v$
$y=vx$
$y=vx+v$
We get
$\frac{dv}{dx}x+v=\frac{2v}{2+{v}^{2}}$
Then
$\frac{dv}{dx}x=-\frac{{v}^{3}}{2+{v}^{2}}$
$\frac{dx}{x}=-\frac{2+{v}^{2}}{{v}^{3}}dv$
$\frac{dx}{x}=\left(-\frac{2}{{v}^{3}}-\frac{1}{v}\right)dv$
Upon integration we have:
$\mathrm{log}x+C=\frac{1}{{v}^{2}}-\mathrm{log}v$
Lets

Elaine Verrett
Rewrite equation into form :
$\frac{dy}{dx}=\frac{2xy}{2{x}^{2}+{y}^{2}}$
Substitute:
$z=\frac{y}{x}⇒y=xz+z$
Therefore:
$xz+z=frac2z2+{z}^{2}Rightarrowxz=\frac{-{z}^{3}}{2+{z}^{2}}⇒\int \frac{2+{z}^{2}}{{z}^{3}}dz=-\int \frac{dx}{x}$