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Differential Equations: Picard's Existence Theorem

  1. Mar 30, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex] y y\prime = 3 [/tex]
    [tex] y(2) = 0[/tex]

    2. Relevant equations

    Solve and find two different solutions.

    3. The attempt at a solution

    [tex] F = \frac{3}{y} [/tex]
    [tex] \frac{\partial F}{\partial y} = \frac{-3}{y^2} [/tex]

    Where do I go from here?
  2. jcsd
  3. Mar 30, 2008 #2
    The problem is a so called separable diff. equation, so write it as
    and integrate on both sides.
    I think you now can proceed by yourself.
  4. Mar 30, 2008 #3


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    Staff Emeritus
    Science Advisor

    Where did you get those? What is F?
    It should be easy to write
    [tex]y\frac{dy}{dt}= 3[/itex]
    [tex]ydy= 3dt[/itex]
    That's basically what eys_physics said.
  5. Mar 30, 2008 #4
    Thanks guys.

    That problem was so simple. I can't believe I didn't see how to do it. ;)
  6. Mar 30, 2008 #5
    To find one solution:
    [tex] \int y dy = \int 3 dx [/tex]
    [tex] y^2 = 6x + C [/tex]
    [tex] y = \pm \sqrt{6x + C} [/tex]

    So given y(2) = 0 I find that C = -12.

    So my two solutions are
    [tex] y = \sqrt{6x + C} [/tex]
    [tex] y = - \sqrt{6x + C} [/tex]

    Last edited: Mar 30, 2008
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