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Euler-Lagrange Equations

  1. Nov 14, 2013 #1
    1. The problem statement, all variables and given/known data

    Compute the Euler-Lagrange for:

    ∫y(y')2+y2sinx dx

    2. Relevant equations

    [itex]\frac{∂L}{∂y}[/itex]-[itex]\frac{d}{dx}[/itex] ([itex]\frac{∂L}{∂y'}[/itex])

    3. The attempt at a solution
    Usual computation by hand gives me y'2+2ysin(x) - 2yy'', but Mathematica says it's -y'2-2yy''. Am I doing something wrong?
  2. jcsd
  3. Nov 14, 2013 #2

    Ray Vickson

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    I get ##L_y - D_x L_{y'} = 2 y \sin(x) - y'^2 - 2 y y''##.
  4. Nov 14, 2013 #3
    Hmmm I looked it over again and I get y'2+2ysin(x)-2(yy''+y'2)
    For the first part I get: y'2+2ysinx
    Second partial: 2yy'
    Then derivative of thatis : 2(yy''+y'2)

    Ahh so it's the same as you got ty. Why there is no sin in mathematica solution T.T
    Last edited: Nov 14, 2013
  5. Nov 15, 2013 #4

    Ray Vickson

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    I don't use or have access to Mathematica (I use Maple instead), so I have no idea what instructions you entered and what Mathematica did with them.
  6. Nov 15, 2013 #5
    Regardless, I assume that's 2ysin(x)−y′2−2yy′ correct Euler-Lagrange then?
  7. Nov 15, 2013 #6


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    Yes it is.
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