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Initial value problem

  1. Dec 11, 2006 #1
    The problem statement, all variables and given/known data
    **********************************************

    Find the two solutions of the initial value problem

    dy/dx = 3y^(2/3) ; y(0) = 0

    The attempt at a solution
    *********************

    I integrate to find the general solution

    y = (x + C)^3

    c=0 is obviously a solution, but I can't find the other one. The clue here is to find a value for c that fits into the equation 0 = c^3, right?
     
    Last edited: Dec 11, 2006
  2. jcsd
  3. Dec 11, 2006 #2

    J77

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    I've got a massive cold today - and my brain's all fuzzy - but I think you should check your integration again...
     
  4. Dec 11, 2006 #3
    It should be right now. Still can't find the solution.
     
  5. Dec 11, 2006 #4

    cristo

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    How have you integrated the equation? Have you separated the equation into the form?
    [tex]\int f(y) dy = \int g(x) dx [/tex]

    If not, I suggest doing so, and posting your attempts
     
  6. Dec 11, 2006 #5

    arildno

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    Dearly Missed

    This diff.eq violates the conditions required in the uniqueness theorem, and has, in fact, an infinite number of solutions.
     
  7. Dec 11, 2006 #6
    What is the uniqueness theorem about? I think it's explained in a bad way in my book. Du kan forklare på norsk hvis du vil.
     
    Last edited: Dec 11, 2006
  8. Dec 11, 2006 #7

    HallsofIvy

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    Uh, it's about uniqueness? If your book doesn't mention that, yes, it's "explained in a bad way"!:rolleyes: Essentially, the "fundamental existence and uniqueness theorem" for initial value problems says that if f(x,y) is continuous and "Lipshchitz" (many text books use the simpler, "sufficient but not necessary" condition "continuously differentiable") on some neighborhood of (x0,y0), then the differential equation dy/dx= f(x,y) has a unique solution satisfying y(x0)= y0.

    It is possible to show, but isn't normally done in elementary courses, that "continuous" is sufficient to show existence of a solution, "Lipschitz" is necessary for "uniqueness".

    Here, dy/dx= 3y2/3. Then y-2/3dy= 3dx. Integrating that, 3y1/3= 3x+ C' or y1/3= x+ C (C= C'/3.) And so y= (x+C)3 just as you say. In particular, if you take C= 0 then y(0)= 0 so y= x3 satisfies your differential equation. However, since (2y2/3)y= (1/3)y-1/3 does not exist at y= 0 so f(x,y) is neither differentiable nor Lipschitz in a neighborhood of (0,0) and that solution may not be "unique".

    In fact, there is an obvious, very simple function that satisfies dy/dx= y2/3, y(0)= 0! (What is the derivative of a constant function?)
     
    Last edited: Dec 11, 2006
  9. Dec 11, 2006 #8
    Ah, I see. So y(x) = 0 is a solution. I guess that must be figured out by inspection.

    But are there more than these two solutions?

    And with "Lipshchitz" you mean that the second derivative is differentiable?
     
    Last edited: Dec 11, 2006
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