I am trying to solve the equation
$$
(x^2-y^2)y' – 2xy = 0.
$$
I have rearranged to get
$$
y' = f(x,y)
$$
where
$$
f(x,y) = \frac{2xy}{x^2-y^2}.
$$
From here I tried to use a trick that I learned from class where we set $x=\frac{y}{x}$ and write
$$
\frac{z'}{f(1,z)-z} = \frac{1}{x}.
$$
After doing a calculation I have that
$$
\frac{z'}{f(1,z)-z} = z'\frac{(1-z^2)}{(z^3+z)}.
$$
From here I would like to use the fact that
$$
z'\frac{(1-z^2)}{(z^3+z)} = \frac{1}{x}
$$
and integrate both sides. However, I'm stuck at integrating the left-hand side.
How can I integrate the left-hand side, or is there is a better way to solve this equation? Thanks.
Best Answer
Correction (after missing a sign:) As kobe pointed out, the original DE is $$ (x^2-y^2)y'-2xy=0, $$ which as equation for a vector field reads $$ (x^2-y^2)\,dy-2xy\,dx=0\iff Im(\bar z^2\,dz)=0\text{ with } z=x+iy. $$ From the complex interpretation it is directly visible that this is not integrable, for that it would have to be an expression $Im(f(z)\,dz)$. But since $\bar z=|z|^2/z$, an integrating factor presents itself as $|z|^{-4}=(x^2+y^2)^{-2}$ which repairs this deficiency. Then $$ \frac{(x^2-y^2)\,dy-2xy\,dx}{(x^2+y^2)^2}=Im(z^{-2}\,dz)=-d(Im(z^{-1})) $$ which implies that all solution trajectories lie on the curves $$ -Im(z^{-1})=\frac{y}{x^2+y^2}=C\quad(\in \Bbb R) $$ which can be solved as $$ y^2-2y(2C)^{-1}+(2C)^{-2}=(2C)^{-2}-x^2\\ y=(2C)^{-1}\pm\sqrt{(2C)^{-2}-x^2} $$ giving the solutions $y\equiv0$ and $y=D\pm\sqrt{D^2-x^2}$ on the interval $x\in[-D,D]$.
Variant: If one reads too fast that the sign before $2xy$ were reversed, then $$ (x^2-y^2)dy+2xydx=0\quad (\iff Im(z^2·dz)=0) $$ is an exact differential equation with conserved quantity or first integral $x^2y-\frac13y^3$ ($=Im(\frac13 z^3)$) so that the solutions lie on the curves $$ y(y^2-3x^2)=const. $$