1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook