1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook