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Here is the problem:

For what values of $\alpha$ and $\beta$, the function

$$\mu(x,y)=x^{\alpha}y^{\beta}$$

is an integrating factor for the OE $$ydx+x(1-3x^2y^2)dy=0.$$ I am working on it just knowing the definition. :(

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2 Answers 2

up vote 2 down vote accepted

Let $$M(x,y)dx+N(x,y)dy=0$$

If exists $F$ such that $$dF=M(x,y)dx+N(x,y)dy$$

then

$$ F(x,y)=C$$

for some $C \in \mathbb{R}$ defines an implicit solution of the ODE

When does $F$ exists? Exists if

$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} $$

and $$M(x,y)=\frac{\partial F}{\partial x}, N(x,y)=\frac{\partial F}{\partial y} $$

Multiply your equation by $\mu(x,y)=x^{\alpha}y^{\beta}$

$$ x^{\alpha}y^{\beta+1}dx+x^{\alpha+1}y^{\beta}(1-3x^2y^2)dy=0$$

Identify $M(x,y)$ and $N(x,y)$ and apply the above theory. I found that $\alpha=\beta=-3$

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Just a good point:

If your OE is of form $$yf(xy)dx+xg(xy)dy=0$$ then you can assume that the integrating factor is $$\mu(x,y)=\frac{1}{xy(f-g)}$$ It means you have it here $$\mu=\frac{1}{xy(1-1+3x^2y^2)}$$

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And a good point at that! +1 –  amWhy May 20 '13 at 0:55
    
@amWhy: Thanks Amy. This is a very useful facts which I gave it to the OP. –  Babak S. May 20 '13 at 16:58

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