I dont see how to make this into an I.F. format:

(3x^2y + 2xy + y^3)dx + (x^2 + y^2)dy = 0

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- Feb 21st 2011, 10:11 PMdmbocciIntegrating factor
I dont see how to make this into an I.F. format:

(3x^2y + 2xy + y^3)dx + (x^2 + y^2)dy = 0 - Feb 21st 2011, 10:23 PMFernandoRevilla
Your equation $\displaystyle Pdx+Qdy$ satisfies:

$\displaystyle \dfrac{1}{Q}\left( \dfrac{\partial P}{\partial y}-\dfrac{\partial Q}{\partial x}\right)=3$

This means that the equation has an integrating factor that depends on $\displaystyle x$ .

Fernando Revilla - Feb 21st 2011, 10:28 PMdmbocci
I came up with 3, although I thought that I was looking for an x or some variable as an answer. How does that show that it depends on x. I dont quite understand.

- Feb 21st 2011, 10:36 PMFernandoRevilla

$\displaystyle 3=3+0x$

Fernando Revilla - Feb 21st 2011, 10:37 PMdmbocci
Nope, still dont get it. heh sorry

- Feb 21st 2011, 10:48 PMFernandoRevilla

$\displaystyle \dfrac{\mu'(x)}{\mu (x)}=3 \Rightarrow\log |\mu(x)|=\int 3\; dx\Rightarrow \ldots$

Fernando Revilla