General solutions to ODE $(x^2-y^2) \frac {dy}{dx}=2xy$ Find the general solutions to ODE
$$(x^2-y^2) \frac {dy}{dx}=2xy$$
Hmm I'm stuck on this one, I thought it was supposed to be solved by Bernoulli's Equation, but it can't.
I've tried to change the equation to various forms.
$$(x+y)(x-y)\frac {dy}{dx}=2xy$$
or
$$\frac {dy}{dx} - \frac{y^2}{x^2} \frac {dy}{dx} = \frac {2y} x$$
But none works.
Please kindly some me some lights. Thanks.
 A: Hint: Make the substitution  $y=vx $. Then the differential equation becomes  $$v+x \frac {dv}{dx}=\frac {2vx^2}{x^2-(vx)^2}=\frac {2v}{1-v^2}.$$So $$x\frac {dv}{dx}=\frac {v+v^3}{1-v^2}.$$ Can you take it from here? 
A: $$(x^2-y^2) \frac {dy}{dx}=2xy$$
$$(2xy)dx+(y^2-x^2)dy=0$$
This is a non-exact differential. To make it exact, one have to find an integrating factor, say $y^{-2}$ in the present case.
$$y^{-2}(2xy)dx+y^{-2}(y^2-x^2)dy=0$$
$$2\frac{x}{y}dx+(1-\frac{x^2}{y^2})dy=0$$
Now this is an exact differential.
$$d\left(y+\frac{x^2}{y} \right)=0$$
$$y+\frac{x^2}{y} =c$$
$$y^2-cy+x^2=0$$
Solve it for $y$ .
A: I post a second answer in order to show how to transform the original ODE into a Bernoulli ODE such as expected by the OP.
$$(x^2-y^2) \frac {dy}{dx}=2xy$$
$$\frac {dy}{dx}=\frac{2xy}{x^2-y^2}$$
In fact, instead of looking for $y(x)$ we will look for the inverse function $x(y)$.
In order to make it more obvious, we change the symbols :
$$\begin{cases}x=Y\\y=X\end{cases}$$
$$\frac {dY}{dX}=\frac {dx}{dy}=\frac{x^2-y^2}{2xy}=\frac{Y^2-X^2}{2YX}$$
$$\frac {dY}{dX}-\frac{1}{2X}Y=-\frac{X}{2}Y^{-1}$$
This is a Bernoulli ODE on the standard form $\quad Y'+p(X)Y=q(X)Y^n$
where $\quad p(X)=-1/(2X)\quad;\quad q(X)=-X/2 \quad;\quad n=-1$.
Then you can proceed on the usual way to solve the Bernoulli ODEs.
