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Long Division to Integrate

  1. Feb 15, 2012 #1
    1. The problem statement, all variables and given/known data
    The problem is about solving the homogenous differential equation (x2 + y2)dx + (x2 - xy)dy = 0 using substitution, in this case y=ux. This is the example they go through in the textbook (A First Course in Differentia Equations with Modeling Applications, 9th Edition, Zill, Example 1 Section 2.5). Anyway, that stuff does not really matter.

    You reach this equation and need to integrate it: [(1 - u)/(1+u)]du + dx/x = 0
    The next equation is [1 - (2/(1+u))]du + dx/x, and it says they got there using long division. I cannot figure out how to go from [(1 - u)/(1+u)]du to [1 - 2/(1+u)]du using long division, since I never learned this, and was wondering if someone could provide a quick explanation on how to do this.


    2. Relevant equations
    Go from [(1 - u)/(1+u)]du to [1 - (2/(1+u))]du using long division (the function needs to be integrated, which is the reason for the long division)


    3. The attempt at a solution

    Any help would be greatly, appreciated. Thanks guys
     
  2. jcsd
  3. Feb 15, 2012 #2

    HallsofIvy

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    You are dividing -u+ 1 by u+ 1. Clearly, u divides into -u -1 times. Multiplying u+ 1 by -1 gives -u- 1 and subtracting the remainder is (-u+ 1)-(-u- 1)= 2.

    That is, u+ 1 divides into -u+ 1 -1 times with remainder 2.

    [tex]\frac{-u+1}{u+1}= -1+ \frac{2}{u+1}[/tex]
     
  4. Feb 15, 2012 #3

    tiny-tim

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    hi solomon684! :smile:
    the best way to explain long division is to actually do it …

    eg (1 + u + 3u2)/(1+u)

    write the top line first, with the highest powers on the left: 3u2 + u + 1

    now look at the first item, it looks like the start of 3u(1 + u)

    so subtract 3u(1 + u) from the whole thing …

    that leaves -2u + 1

    now that looks like the start of -2(1 + u)

    so subtract -2(1 + u) from the whole thing …

    that leaves 3

    so the answer is: (3u2 + u + 1)/(1 + u) = 3u - 2 + 3/(1 + u) :smile:
     
  5. Feb 15, 2012 #4
    Hm, I kind of get it, but at the same time I'm still pretty lost and feel like I'd never be able to figure it out on my own.

    I looked up polynomial long division and tried doing it the way the first example here is done, and got an answer of -1 with remainder 2. When doing it that way, is it safe to say it can always be rewritten as answer + remainder/original denominator? Because that is how this example works out but I am not sure if it is always true
     
  6. Feb 15, 2012 #5

    Ray Vickson

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    Whether or not you call it long division, it IS the type of thing you must have done in Calculus 101, when you took integration. [tex] \frac{1-u}{1+u} = \frac{2-(1+u)}{1+u} = \frac{2}{1+u} - 1. [/tex] So, I don't go from (1-u)/(1+u) to 1 - 2/(1+u), but I do go to 2/(1+u) - 1.

    RGV
     
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