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Homework Help: Calculus of Variations - more confusion

  1. Jan 22, 2009 #1
    I asked a question earlier about Calculus of Variation, but the question I gave didn't really highlight my confusion well. I've come across some other questions that I think reveal my misunderstanding.

    1. The problem statement, all variables and given/known data

    Solve the Euler equation for the following integral:

    (integral from x1->x2) ∫[(y')² + √y]dx

    2. Relevant equations

    Euler equation:

    ∂F/∂y - d/dx (∂F/∂y') = 0

    3. The attempt at a solution

    F = (y')² + √y

    So ∂F/∂y = 1/[2√y] and d/dx (∂F/∂y') = 2y'

    Thus: y'' = 1/[4√y]

    Although it has been some time since I took an ODE course, I think that the equation above is non-trivial to solve. So either I'm mistaken and this is easy to solve or I'm going about the calculus of variations method mistakenly.

    With a similar problem, ∫[1+yy']²dx, I was only able to reduce it down to: y''y² + y(y')² = 0, which again I couldn't solve.

    I'm not looking for answers, just want to know where I'm applying the method incorrectly, or if in fact I'm missing a far easier way to apply it.

    Thanks for any help!
     
  2. jcsd
  3. Jan 22, 2009 #2
    Any thoughts?

    Is y'' = 1/[4√y] easily solvable? If not how else can the method be applied?

    Cheers
     
  4. Jan 22, 2009 #3

    Dick

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    As far as calculus of variations goes, they look fine. Just because you get a difficult ODE doesn't mean the method is wrong.
     
  5. Jan 22, 2009 #4

    gabbagabbahey

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    Try solving for y' first by noting that

    [tex]y''(x)=\frac{d}{dx}y'(x)=\left(\frac{d}{dy}\frac{dy}{dx}\right)y'(x)=\frac{dy'}{dy}y'[/tex]
     
    Last edited: Jan 22, 2009
  6. Jan 22, 2009 #5
    Thanks gabbagabbahey, that was a big help.
     
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