Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$

Question

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 $3yx^2 - y^3 = k$

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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!!!

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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.

Answer 1

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?

Answer 2

$$y' = \frac{2xy}{y^2-x^2}$$ Since $y'=\dfrac 1 {x'}$: $$\dfrac 1 {x'} = \frac{2xy}{y^2-x^2}$$ $${y^2-x^2} = {2xx'y}$$ Note that $2xx'=(x^2)'$: $$y^2 = {(x^2)'y}+x^2$$ $$y^2 = {(x^2y)'}$$ Integrate: $$\dfrac {y^3}3=x^2y+C$$