# Integrating Factor

• Jan 15th 2011, 05:49 PM
ragnar
Integrating Factor
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.
• Jan 15th 2011, 08:21 PM
Jester
This is what I see - I don't know if it'll help

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

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

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

$\displaystyle \dfrac{d \left( \dfrac{x}{y} \right)}{\left(\dfrac{x}{y}\right)^2 + 1 } - dy = 0$.
• Jan 15th 2011, 08:24 PM
dwsmith
Quote:

Originally Posted by Danny

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

$\displaystyle 2. \ \ \ d \left( \dfrac{x}{y} \right)$

How can you go from 1 to 2?
• Jan 15th 2011, 08:38 PM
ragnar
Quote:

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

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

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

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

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

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!
• Jan 15th 2011, 09:27 PM
Random Variable
Quote:

Originally Posted by dwsmith
How can you go from 1 to 2?

quotient rule in reverse
• Jan 15th 2011, 09:32 PM
Random Variable
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.
• Jan 16th 2011, 06:32 AM
Jester
Quote:

Originally Posted by 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.

Correct me if I'm wrong but isn't one missing from here. If the integrating factor is a function of $\displaystyle x^2+y^2$ then

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