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Easy diff. eq. problem

  1. Sep 27, 2005 #1
    Given: y=(x-c)^2 is a solution to
    (y')^2=4y

    If y(x0)=y0, is that the ONLY solution?
    I'm not terribly clear on the existence and uniqueness theorem(the technical math aspect isn't so bad for me, it's the stuff like that)so fill in where I'm wrong!

    so for the exist. and uniq. theorem to apply, y'=f(x,y), f(x,y) and the partial of f in terms of y must be continuous.

    Well, the function then is 2y^(1/2). It's only continuous when y >or= 0, and the partial is y^(-1/2) which is only continuous when y>0. So the function isn't guaranteed to have a unique solution unless y0>0, right?

    so to further learn I solved it, and got for my answer, y=(x+c)^2. Which is basically the same thing as the given solution(since c could be taken as positive or negative)

    so, uh, how do I intepret that? Is it not guaranteed an unique solution but happens to only have one or did I somehow not find one?
     
  2. jcsd
  3. Sep 27, 2005 #2

    saltydog

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    For the equation:

    [tex](y^{'})^2=4y\tag{1}[/tex]

    Solving for [itex]y^{'}[/itex], I'd take the square root of both sides? Two equations result when that happens right? Solutions to each of these equations will then satisfy the original equation. What does that tell you about uniqueness?

    Suppose I had to calculate what c would have to be in order for the solution of (1) to pass through the point ([itex]x_0,y_0[/itex]).

    Then I'd need to solve:

    [tex]y_0=(x_0-c)^2[/tex]

    for c. Taking square roots of both sides and not forgetting the plus and minus part, What can c be in this case?

    The existence and uniqueness theorem you mention applies to equations of the form:

    [tex]y^{'}=f(x,y)[/tex]

    Treating each of the two separate ODEs calculated above separately, the consequence of this theorem applies.
     
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