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**ragnar** Working with Tenenbaum's ODE page 91, problem 17, $\displaystyle y dx - (y^{2} + x^{2} + x)dy = 0$, can't figure it out. Here are some pieces of information that could be useful, given the methods of finding an integrating factor discussed in the chapter:

$\displaystyle \frac{d}{dy}P(x,y) = 1, \frac{d}{dx}Q(x,y) = -2x -1 \Rightarrow$

$\displaystyle \frac{d}{dy}P(x,y) - \frac{d}{dx}Q(x,y) = -2(x+1)$

(a)$\displaystyle yQ(x,y) - xP(x,y) = -y(y^{2} + x^{2} + 2x)$

(b)$\displaystyle yQ(x,y) + xP(x,y) = -y(y^{2} + x^{2})$

Now I need to take $\displaystyle \frac{d}{dy}P(x,y) - \frac{d}{dx}Q(x,y)$ or it times $\displaystyle x^{2}$ or $\displaystyle y^{2}$, and divide it by (a) or (b) so that the result is a function of $\displaystyle x, y, xy, x/y$, or $\displaystyle y/x$. But none of it seems to work. The book says that the integrating factor necessary is $\displaystyle \frac{1}{x^{2} + y^{2}}$, but I don't see how you could possibly obtain that with the methods discussed in the chapter.