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Easy differential equation

  1. Sep 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Solve x.dy/dx+1-y^2=0


    2. Relevant equations



    3. The attempt at a solution

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

    The LHS can be broken into
    dy/(2(y-1))-dy/(2(y+1))

    Integrating:

    Log[y-1]/[y+1] = log[x^2] + c

    Given x=1 when y=0 c=0

    Y-1=x^2.y+x^2

    Y=(x^2+1)/(x^2-1)

    The answer is the reciprocal of mine, can anyone see my mistake?

    Thanks
     
  2. jcsd
  3. Sep 16, 2011 #2

    CompuChip

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    Actually it's not the reciprocal, it is the negative:
    y = (x^2+1)/(1-x^2) = -(x^2+1)/(x^2-1)
    is the correct answer.

    There is a simple algebraic error in the last step, everything is correct up to
    y-1=x^2.y+x^2

    Then,
    y - x^2 y = 1 + x^2
    (1 - x^2) y = (1 + x^2)
    y = (1 + x^2) / (1 - x^2)
     
  4. Sep 16, 2011 #3
    http://www.wolframalpha.com/input/?i=solve+x*dy/dx+++(1-y^2)=0+

    Weirdly, I can't get this question right doing it as above, but if I divide by (1-y^2) it does work.

    If you drop modulus signs, the constant wasn't 0 but 0.5log(-1).

    Not really sure why the mod signs mess it up like that..
     
    Last edited: Sep 16, 2011
  5. Sep 16, 2011 #4

    lurflurf

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    The answer is
    y=(1-x^2)/(1+x^2)
    I think you lost a sign with the logarithms.
    y=(C-x^2)/(C+x^2)
    y=1
    for example are solutions
    Given x=1 when y=0 we see C=1.
    Your answer is the case C=-1 which does not satisfy the initial conditions, but since you were working in absolute value the condition |C|=1 was satisfied.
     
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