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!

Partial Fractions in Differential Equations

  1. Jul 19, 2014 #1
    I'm a little rusty with partial fractions, and I can't seem to find my error once I get up to that point.

    1. The problem statement, all variables and given/known data

    dy/dx = (y^2 - 1) / x

    2. Relevant equations


    3. The attempt at a solution

    Cross-mutliply

    x dy = (y^2 - 1) dx

    Divide by the appropriate terms

    dy / (y^2 - 1) = dx / x

    So I'm integrating 1 / (y^2 - 1) dy and 1/x dy

    Obviously, the integral of 1/x is ln|x| + c, but I'm having trouble with integrating the y terms. This is what I did so far.

    A / (y + 1) + B/ (y - 1) = 1 / (y^2 - 1)

    Ay - A + By + B = 1

    Combine the y terms and constant terms

    A + B = 0 (Since there's no y term)
    -A + B = 1

    2B = 1

    B = 0.5

    So I integrate 0.5 / (y - 1) and get 0.5*ln|y - 1|

    So that means 0.5*ln|y - 1| = 1/|x| + c

    ln|y - 1| = 2/x + c

    y - 1 = e^(2/x + c)

    y - 1 = e^(2/x) * e^c (which is just a constant)

    y = ce^(2/x) + 1

    But the thing is that I checked online to find out how to integrate the y term, and found this: http://calc101.com/partial_1.html

    My question is: Why does 1 / (y + 1)(y - 1) get split up into 1/(2y - 2) and 1/(2y + 2)? Also, why are they subtracting those 2 terms rather than adding them?
     
  2. jcsd
  3. Jul 19, 2014 #2

    verty

    User Avatar
    Homework Helper

    You found that B = 1/2, but you forgot to find a value for A. Also notice that ##{0.5 \over (y - 1)} = {1 \over (2y - 2)}##.
     
  4. Jul 19, 2014 #3
    Wow, I guess I saw that A + B = 0, and somehow thought A was 0.

    So if B is 1/2, that means A = -1/2, so I'm integrating -0.5/(y + 1) + 0.5(y - 1), which gets me -0.5*ln|y + 1| + 0.5*ln|y - 1| = ln|x| + c

    Multiplying both sides by 2 gets me:

    -ln|y + 1| + ln|y - 1| = 2*ln|x| + c

    Raise e to both sides.

    e^(1/(y + 1)) * e^(y - 1) = e^(2*ln|x| + c)

    e^(1/(y + 1)) * e^(y - 1) = e^(2*ln|x|) * e^c

    e^(1/(y + 1)) * e^(y - 1) = ce^(2*ln|x|)

    Is it possible to get that in explicit form? Or should I divide both sides by e^(2ln|x|) and just leave it in terms of c?

    Thank you very much!
     
  5. Jul 20, 2014 #4

    verty

    User Avatar
    Homework Helper

    It is possible to get a closed form solution. That said, it is a little tricky.

    You seemed quite eager to get rid of the logs as soon as possible, but the log form is actually easier to work with. Actually, you made a mistake in your algebra. Perhaps look over the log/exponent laws to refresh your knowledge.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted