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Problem with my ODE

  1. Oct 24, 2007 #1
    I'm having trouble finding the solution to the following ODE:

    y'=(y^3)/2 with initial value y(0) = 1

    I try to separate it but end up with


    which makes no sense, since you can't take the square root of a negative number.

    Any help?

    Lauren. =)
  2. jcsd
  3. Oct 24, 2007 #2
    You *can* take the square root of a negative number, but the answer will be complex. Don't forget your constants of integration also, which should help you solve this problem.
    Last edited: Oct 24, 2007
  4. Oct 24, 2007 #3


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    Even worse 1/x is not defined at x=0. You are forgetting to put in an integration constant.
  5. Oct 24, 2007 #4
    Good point with the constants. So....

    1/(y^3) dy = 1/2 dx Integrate both sides

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

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

    this solves the problem with it being undefined at 0, but at this point, I guess I need to add my complex number i?

    y = i/(x+2c), which would make c=i/2 and so y = 2i/(2x + ci)....I think.

    Is this right?

    we haven't really used complex numbers so far in this particular course, which was why I was hesitant to use it in the answer.

  6. Oct 24, 2007 #5


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    Nooo. Just put y=0 and x=1 and solve for C. Those are your initial conditions. I don't think you have to solve for y.
  7. Oct 24, 2007 #6
    okay....so, then, plugging in my initial values at the Y^2 point, I get:

    1^2=-1/(0+2c) ==> c=-1/2 ==> y^2=-1/(x-1) ==>

    y^2 = 1/(1-x) ==> y = sqrt(1/(1-x))

    Is that it?

  8. Oct 24, 2007 #7


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    Yep, that's it.
  9. Oct 24, 2007 #8
    Thanks....much appreciated. =)
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