1. ## Integrating Factor

Working with Tenenbaum's ODE page 91, problem 17, $y dx - (y^{2} + x^{2} + x)dy = 0$, 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:

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

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

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

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

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

2. This is what I see - I don't know if it'll help

$ydx - x dy - (x^2+y^2)dy = 0$

$\dfrac{ydx-xdy}{y^2} - \left(\dfrac{x^2}{y^2} + 1 \right)dy = 0$

$d \left( \dfrac{x}{y} \right) - \left( \left(\dfrac{x}{y}\right)^2 + 1 \right)dy = 0$

$\dfrac{d \left( \dfrac{x}{y} \right)}{\left(\dfrac{x}{y}\right)^2 + 1 } - dy = 0$.

3. Originally Posted by Danny

$1. \ \ \dfrac{ydx-xdy}{y^2}$

$2. \ \ \ d \left( \dfrac{x}{y} \right)$
How can you go from 1 to 2?

4. Originally Posted by Danny
This is what I see - I don't know if it'll help

$ydx - x dy - (x^2+y^2)dy = 0$

$\dfrac{ydx-xdy}{y^2} - \left(\dfrac{x^2}{y^2} + 1 \right)dy = 0$

$d \left( \dfrac{x}{y} \right) - \left( \left(\dfrac{x}{y}\right)^2 + 1 \right)dy = 0$

$\dfrac{d \left( \dfrac{x}{y} \right)}{\left(\dfrac{x}{y}\right)^2 + 1 } - dy = 0$.
If I understand you, the variables are separated and then next step is to just straight-up integrate like a boss. If that's correct then I doubt it's the intended solution since that uses (1) a cleverness beyond my current capacity, (2) basic principles of the preceding chapter, and (3) none of the techniques discussed in the concurrent chapter.\footnote{Yeah, that's definitely not the right use of that word, but whatev.} But it is good for me to see such a technique for future problems. Thank you!

5. Originally Posted by dwsmith
How can you go from 1 to 2?
quotient rule in reverse

6. The answer suggests that the integrating factor is h(x/y) or h(y/x). But neither of the methods in the book seem to work.

7. Originally Posted by ragnar
Working with Tenenbaum's ODE page 91, problem 17, $y dx - (y^{2} + x^{2} + x)dy = 0$, 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:

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

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

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

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

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

$\dfrac{P_y-Q_x}{yP-xQ}$ is then a fucntion of $x^2+y^2$.