$2xyy' = x^2-y^2$ I have to solve this homogeneous equation (I believe it's homogeneous) and the farther I can get is:


*

*$$2xyy' = x^2-y^2$$

*$$y' = \frac{x}{2y} - \frac{y}{2x}$$


So, from here, I need explanations on how do I get to the form:
$$y' = f(\frac xy)$$
Thanks guys!
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Lets $\ds{x = \alpha\tilde{x}}$ and $\ds{y = \beta\,\tilde{y}}$. Then,
  $$
2xyy'=x^2 - y^{2}\implies
2\tilde{x}\,\tilde{y}\,\tilde{y}' = \pars{\alpha \over \beta}^{2}\tilde{x}^{2} - \tilde{y}^{2}
$$

The equation is invariant under the above scaling whenever $\ds{\alpha = \pm\beta}$. It means,
$$
{x \over \tilde{x}} = \pm\,{y \over \tilde{y}}\implies
{\tilde{y} \over \tilde{x}} = \pm\,{y \over x}
$$
It suggests the change of variables $\ds{u \equiv {y \over x} \implies y = ux}$.
Namely,
\begin{align}
&2x\pars{ux}\pars{u'x + u} = x^{2} - \pars{ux}^{2} \implies
2xuu' + 2u^{2} = 1 - u^{2} \implies
{2u \over 1 - 3u^{2}}\totald{u}{x} = {1 \over x}
\end{align}
Moreover,
\begin{align}
&-\,{1 \over 3}\,\ln\pars{1 - 3u^{2}} = \ln\pars{x} + \ln\pars{\mbox{C}}
\,,\qquad
\pars{~\mbox{C is an integration constant}~}
\\[5mm]
\ln\pars{C} & = \ln\pars{\bracks{1 - 3 u^{2}}^{1/3}x}\implies
C = \pars{1 - 3\,{y^{2} \over x^{2}}}^{1/3}x \implies
\bbx{\ds{y = \pm\,{\root{x^{3} - C^{3}} \over \root{3}x^{1/2}}}}
\end{align}
A: You could treat $2xyy'=x^2-y^2$ as homogeneous (after a bit of algebra). But this is an exact equation:
$$(-x^2+y^2)+2xyy'=0 \qquad \Longrightarrow \qquad (-x^2+y^2)dx+2xy\,dy=0$$
You have $P=-x^2+y^2$ and $Q=2xy$. Notice that $P_y=2y=Q_x$. Thus this is exact.
$\int P\,dx = -\dfrac{1}{3}x^3+xy^2+C_1(y)$ and $\int Q\,dy = xy^2+C_2(x)$. Therefore, $-\dfrac{1}{3}x^3+xy^2=C$ is the general (implicit) solution. Solving for $y$ is easy. :) 
