Integrating Factor

**ragnar** · January 14th 2011, 04:36 PM

I'm doing Tenenbaum's ODE page 91, problem 16, can't see what's going on. We have $3(x+y)^{2}dx + x(3y + 2x)dy = 0 \Rightarrow \frac{d}{dy}P(x,y) = 6(x+y) \text{and} \frac{d}{dx}Q(x,y) = 3y + 4x \Rightarrow$ I want to find an integrating factor by finding one of the forms in which I can divide the difference of these, $2x + 3y$ , by $yQ - xP$ (or there are formulae if this is off by a factor of $-1$ ). Or I want $yQ+xP$ to divide $2x + 3y$ times a factor of $x^{2}$ or $y^{2}$ .

But $yQ - xP = -x^{2}(4y + 3x)$ (grrrrr! so close!

)

And $yQ + xP$ only factors into an $x$ and something bad.

So... I got nothing.

**ragnar** · January 14th 2011, 04:38 PM

GAAAH, I see it now. Nevermind.

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