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ODE y`+2y=y^2

  1. Oct 16, 2005 #1
    for the following question:

    my problem:
    suppose u=1/y
    so u`=-[y^(-2)]*y`=-1+2u
    so du/(1+2u)=-dx
    so y=1/u=2/[1-ce^(-x)]

    but the correct answer should be y=x/[1+2cx^(2x)]
    does anybody know where my calculations went wrong?
  2. jcsd
  3. Oct 16, 2005 #2


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    Homework Helper

    missed out dividing by 2 when integrating.

    so du/(1+2u)=-dx
    (1/2)ln(1+2u) = -x +lnC
    ln(1+2u) = -2x + lnC²
    ln{(1+2u)/C²} = -2x
    1+ 2u = C²e^(-2x)
  4. Oct 16, 2005 #3
    opps~ thanks!!! :)
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