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Math Help - Integrating Factor

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    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.
    Last edited by ragnar; January 15th 2011 at 06:49 PM. Reason: Improve clarity
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    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.
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    Quote Originally Posted by Danny View Post

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

    2. \ \ \ d \left( \dfrac{x}{y} \right)
    How can you go from 1 to 2?
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    Quote Originally Posted by Danny View Post
    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!
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    Quote Originally Posted by dwsmith View Post
    How can you go from 1 to 2?
    quotient rule in reverse
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    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.
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  7. #7
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    Quote Originally Posted by ragnar View Post
    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.
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