Differential equations: $f(x,y) dx + g(x,y) dy = 0$

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

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

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 $y\left(1\right)$.