<s>[INST] Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$ Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$
substituting $y=ux$ so that $y' = u + x\frac{du}{dx}$"
$u + x\frac{du}{dx} = \frac{2x^2u}{ux^2-x^2} = \frac{2u}{u^2-1}$
$\rightarrow x\frac{du}{dx} = \frac{2u}{u^2-1} - u = \frac{2u}{u^2-1} - \frac{u(u^2-1)}{u^2-1}$
$\rightarrow x\frac{du}{dx} = \frac{-u^3+u}{u^2-1}$
Separating variables...
$\rightarrow \frac{u^2-1}{-u^3+u}du = \frac{1}{x}dx$
Now, from here we can factor the top and bottom of the LHS as so...
$\frac{u^2-1}{-u^3+u} = \frac{(u+1)(u-1)}{-u(u+1)(u-1)} = \frac{-1}{u}$
And I'm thinking this is where I'm going wrong... Is there a reason that this simplification is not valid? I'm thinking that maybe I shouldn't simplify and that I should use partial fraction's instead?
The answer is supposed to be $3yx^2 - y^3 = k$

Edit: I continued as so:
$\frac{u^2-1}{3u-u^3} = \frac{A}{u} + \frac{B}{u-\sqrt{3}} + \frac{C}{u + \sqrt{3}}$.
I got $A = \frac{1}{3}$ by multiplying both sides by $u$ and then taking the limit as $u \rightarrow 0$. A similar method led to $B = \frac{1}{3}$ and $C = \frac{1}{3}$. Thus I arrived at:
$\frac{u^2-1}{3u-u^3} = \frac{1}{3u} + \frac{1}{3(u-\sqrt{3})} + \frac{1}{3(u+\sqrt{3})}$.
And so:
$(\frac{1}{3u} + \frac{1}{3(u-\sqrt{3})} + \frac{1}{3(u+\sqrt{3})})du = \frac{1}{x}dx$
Integrating and then exponentiating both sides:
$(u(u+\sqrt{3})(u-\sqrt{3}))^{\frac{1}{3}} = xk$
$(u(u^2-3))^{\frac{1}{3}} = xk$
$u = \frac{y}{x}$ and so:
$(\frac{y}{x}((\frac{y}{x})^2-3))^{\frac{1}{3}} = xk$
This leads to:
$(\frac{y}{x})^3 - \frac{3y}{x} = x^3k$
... Which is still not correct!!!

I haven't done ODE's quite a long time... I've been flying through Schaum's outline on the topic and I've done plenty of problems like this one... But i can't get this one correct!! help appreciated.
 [/INST] HINT
According to the substitution $y = ux$, it results that
\begin{align*}
y' = \frac{2xy}{y^{2} - x^{2}} & \Longleftrightarrow y' =\frac{2(y/x)}{(y/x)^{2} - 1}\\\\
& \Longleftrightarrow xu' + u = \frac{2u}{u^{2} - 1}\\\\
& \Longleftrightarrow xu' = \frac{2u}{u^{2} - 1} - u\\\\
& \Longleftrightarrow xu' = \frac{3u - u^{3}}{u^{2} - 1}
\end{align*}
Can you take it from here?
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