The correct Euler-Lagrange equation for the functional ∫y(y')² + y²sin(x) dx is derived as 2y sin(x) - y'² - 2yy''. The computation involves applying the Euler-Lagrange formula, which is ∂L/∂y - d/dx(∂L/∂y'). Users reported discrepancies between their manual calculations and results from Mathematica, with some obtaining -y'² - 2yy'' instead. The consensus confirms that the correct formulation does include the term 2y sin(x).
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~Sam~ said:Homework Statement
Compute the Euler-Lagrange for:
∫y(y')2+y2sinx dx
Homework Equations
\frac{∂L}{∂y}-\frac{d}{dx} (\frac{∂L}{∂y'})
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?
Ray Vickson said:I get ##L_y - D_x L_{y'} = 2 y \sin(x) - y'^2 - 2 y y''##.
~Sam~ said: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
Ray Vickson said: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.
~Sam~ said:Regardless, I assume that's 2ysin(x)−y′2−2yy′ correct Euler-Lagrange then?