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

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

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

So... I got nothing.