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Please tell me what I did wrong when solving this DE

  1. Sep 3, 2013 #1
    Hello everyone,

    I have a brief question. I am trying to solve the following simple differential equation:
    [itex] x\dfrac{dy}{dx}= y^2-1[/itex]
    I manage to solve it one way, but when I try to solve it in the following way, I cannot get the correct answer. I would really appreciate if someone could point out my mistake :)

    [itex] \dfrac{dy}{y^2-1}= \dfrac{dx}{x}[/itex]
    Rewriting the fraction on the left as the sum of two fractions
    [itex] \dfrac{1}{2}(\dfrac{-1}{y+1}+\dfrac{1}{y-1})dy= \dfrac{dx}{x}[/itex]
    Integrating both sides:
    [itex] \dfrac{1}{2}∫(\dfrac{-1}{y+1}+\dfrac{1}{y-1})dy= ∫\dfrac{dx}{x}[/itex]
    [itex] \dfrac{1}{2}∫(-1ln(y+1)+ln (y-1))dy= ln(x)+c[/itex]
    [itex] \dfrac{1}{2} ln(\dfrac{y-1}{y+1})= ln(x)+c[/itex]
    It is given that when x=1, y=0, thus c=0
    [itex] \dfrac{1}{2} ln(\dfrac{y-1}{y+1})= ln(x)[/itex]
    [itex] \dfrac{y-1}{y+1}= x^2[/itex]
    Which gives:
    [itex] y= \dfrac{1+x^2}{1-x^2}[/itex]
    But, he correct result is:
    [itex] y= \dfrac{1-x^2}{1+x^2}[/itex]

    I would really appreciate if anyone could point out the mistake. Thank you. :)
     
  2. jcsd
  3. Sep 3, 2013 #2
    Hi asdfghhjkl :smile:

    Welcome to Physicsforums !!!

    [itex]\dfrac{y-1}{y+1}= x^2[/itex] should be [itex]\dfrac{|y-1|}{|y+1|}= x^2[/itex]

    and this will give you [itex]\dfrac{1-y}{y+1}= x^2[/itex] .You can check that from the initial conditions.

    Remember ∫(1/x)dx =ln|x|.
     
  4. Sep 3, 2013 #3
    Oh, I see thank you very much for your help :)
     
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