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How do I solve this differential equation? $$y(2x+y^2)dx+x(y^2-x)dy=0$$

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closed as off-topic by alexqwx, Mark Fantini, Davide Giraudo, user133281, Jean-Claude Arbaut Aug 31 at 14:30

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Integrating Factor. –  Nameless Aug 31 at 4:58

1 Answer 1

By dividing the equation by $y^2dx$ and reordering the terms, the equation can be written in the following form: $$(y + xy')+\left(\frac{2x}{y}-\frac{x^2 y'}{y^2}\right) = 0\tag{1}$$ Integrating both sides of the equation with respect to $x$ gives: $$ xy + \frac{x^2}{y} = K \tag{2}$$ Which is a quadratic equation with the following solution: $$ y = \frac{-K\pm\sqrt{K^2-4x^3}}{2x}.\tag{3}$$

$\hspace2in$enter image description here $\hspace2in$A plot of the solutions for $K\in\{-2,-1,0,1,2\}.$

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-1 There is no explanation, this will not help anyone who doesn't know how to do this already. –  eBusiness Aug 31 at 9:55
@eBusiness: Honestly I don't know what kind of explanation to add, I just played a bit with the original differential equations till getting something that looked integrable. $(1)\Rightarrow(2)$ is straightforward, and $(2)\Rightarrow(3)$ is straightforward, too. –  Jack D'Aurizio Aug 31 at 10:00
What kind of transformations did you use? Perhaps with links to relevant articles. The only one that doesn't look totally arbitrary to me is the final transformation. –  eBusiness Aug 31 at 10:11
@eBusiness: To get $(1)$, I just divided the original differential equation by $y^2 dx$, no transformations at all! –  Jack D'Aurizio Aug 31 at 10:16
That is a transformation. –  eBusiness Aug 31 at 10:19

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