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What I'm trying to solve is

$$ \frac{dx}{dy}=\frac{-x- y^3}{y - x^3}.$$

I've tried with substituting with $ v = \frac{x}{y} $ and $ u = \frac{y}{x},$ but even with that I still can't separate the variables.

Using the second substitution I get $$ ( u + x^3u^3 - x^2 + x ) dx + (u + u^4x^2) du = 0$$

I tried working with this but found no way to simplify or advance from this in any way. I plugged it in Wolfram Alpha and of course it gave a nice looking solution to this (an ugly one for the first form of the equation though), so I know I'm missing something.

Any help would be greatly appreciated.

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  • $\begingroup$ That ODE doesn't look very homogeneous to me... $\endgroup$ Jun 23 '11 at 16:57
  • $\begingroup$ @Hans you're right actually, I wasn't really thinking when I wrote the title. $\endgroup$
    – Bananas
    Jun 23 '11 at 17:15
  • $\begingroup$ You should try to find an integrating factor: en.wikipedia.org/wiki/Integrating_factor $\endgroup$ Jun 23 '11 at 17:17
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To complement Jack's answer, you might want to look at this on how to proceed.
All the best.

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  • $\begingroup$ I tried to find similar problems posted here but must have skipped this one, since I didn't know I was looking for the exact equation method. Thanks! $\endgroup$
    – Bananas
    Jun 23 '11 at 19:18
  • $\begingroup$ You are welcome...glad to help..:) $\endgroup$
    – Nana
    Jun 23 '11 at 19:19
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Rewrite it as $$(y-x^3)dx+(x+y^3)dy=0$$ and note that $$\frac{\partial}{\partial y}(y-x^3)=\frac{\partial}{\partial x}(x+y^3)$$ Then you can go on yourself.


According to the substitution you used: $v=\frac{x}{y}$, I guess you were thinking about the homogeneous first-order differential equation which can be written in the form $$ \frac{dy}{dx}=F(\frac{y}{x})$$

This is different from the equation in your question.

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  • $\begingroup$ I'm not too sure how to go from here, but at least I see what I should be doing. I don't have my books right now with me, so when I get home I'll take a look at what Javier suggested and try to solve this. $\endgroup$
    – Bananas
    Jun 23 '11 at 17:36

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