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$$(2xy + 3y^2)dx + (x^2 + 6xy - 2y)dy = 0$$ $$y(1) = -1/2$$

How do you solve this? I have just started learning Differential equations and I have some trouble.

Is this equivalent with this?

$$2y + (6x - 2) = 0$$ $$y(1) = -1/2$$

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    See [exact equations](http://en.wikipedia.org/wiki/Exact_differential_equation).2012-12-12
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    Notice that $(2xy+3y^2)dx+(x^2+6xy-2y)dy$ is the divergent of $F(x,y)=x^2y+3xy^2-y^2+c$, c is a constant.2012-12-12
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    @GustavoMarra I think you mean it is $d{\bf r} \bullet \nabla F$.2012-12-12
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    ooops, that is correct, sorry!2012-12-12
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    See [here](http://math.stackexchange.com/questions/215324/proof-for-exact-differential-equations-shortcut/215346#215346).2012-12-13

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You can rewrite your equation as $$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy = 0, $$ where $df$ is the total differential of a function $f\left(x,y\right)$ that you should determine. Then $df = 0$ implies $f\left(x,y\right)$ is a constant. Determine this constant with the condition on $y\left(1\right)$.