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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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    $\begingroup$ See exact equations. $\endgroup$ – Artem Dec 12 '12 at 22:45
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    $\begingroup$ 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. $\endgroup$ – Marra Dec 12 '12 at 22:52
  • $\begingroup$ @GustavoMarra I think you mean it is $d{\bf r} \bullet \nabla F$. $\endgroup$ – Eric Angle Dec 12 '12 at 23:14
  • $\begingroup$ ooops, that is correct, sorry! $\endgroup$ – Marra Dec 12 '12 at 23:16
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    $\begingroup$ See here. $\endgroup$ – Mhenni Benghorbal Dec 13 '12 at 0:31
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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 $y\left(1\right)$.

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