# How do I Solve This Kind of Differential Equation? [closed]

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

-

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – alexqwx, Mark Fantini, Davide Giraudo, user133281, Jean-Claude Arbaut
If this question can be reworded to fit the rules in the help center, please edit the question.

Integrating Factor. – Nameless Aug 31 '14 at 4:58

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: $$y = \frac{-K\pm\sqrt{K^2-4x^3}}{2x}.\tag{3}$$
$\hspace2in$ $\hspace2in$A plot of the solutions for $K\in\{-2,-1,0,1,2\}.$
@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 '14 at 10:00
@eBusiness: To get $(1)$, I just divided the original differential equation by $y^2 dx$, no transformations at all! – Jack D'Aurizio Aug 31 '14 at 10:16