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Homework Help: Is partial fraction wrong way to go?

  1. Dec 20, 2006 #1
    1. The problem statement, all variables and given/known data
    Solve the separable equation


    2. Relevant equations
    dy/dx = 1-y^2


    3. The attempt at a solution
    dy/dx = 1-y^2

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

    [Partial fraction]
    A/(1-y) + B/(1+y) = 1/(1-y^2)

    A + Ay + B - By = 1
    y^1: A - B = 0
    y^0: A + B = 1
    => A=B=1/2 =>

    (1/2)/(1-y) dy + (1/2)/(1+y) dy = dx

    ln|1-y| + ln |1+y| = 2x + C

    ln|1-y^2| = 2x + C

    1-y^2 = De^(2x)

    y = sqrt(1 - De^(2x))

    This answer is wrong according to two different books (without explanation) that i have. The correct answer should be
    y = (De^(2x) - 1)/(De^(2x) + 1)


    Is partial fraction wrong way to go?
    Have I made a wrong turn along the way with the algebra?
    I do have big problems when i comes to solve nonlinear integrals, a tip along the way would be very appreciated!
     
  2. jcsd
  3. Dec 20, 2006 #2
    Check the second step.
     
  4. Dec 20, 2006 #3
    Of course!

    So obvious now, thanks a lot!!!
     
  5. Dec 20, 2006 #4
    Minus signs are your friends.
     
  6. Dec 21, 2006 #5
    Dun like partial fractions?? Nevermind...

    Hi m3cklo,

    If u dun want to break the integral into partial fractions, u can use a trigo substitution instead. Let y = sin t. Then, dy = cos t dt. Simplify using the substitution. U should get sec t as the integral. Integrate it and u get ln (sec t + tan t). Convert the result back to y. U should get the same answer as the one using partial fractions.

    Hope that helps.
     
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