The discussion revolves around computing the Euler-Lagrange equation for the integral ∫y(y')² + y²sin(x) dx, focusing on the correct formulation and potential discrepancies between manual calculations and software outputs.
There is ongoing dialogue about the calculations, with participants reviewing their work and discussing the outputs from different software. Some guidance is provided in terms of confirming the correctness of the derived expressions, but no consensus on the final form has been reached.
Participants mention using different computational tools (Mathematica vs. Maple), which may influence the results and interpretations of the Euler-Lagrange equation. There is also a noted uncertainty regarding the absence of certain terms in the Mathematica output.
~Sam~ said:Homework Statement
Compute the Euler-Lagrange for:
∫y(y')2+y2sinx dx
Homework Equations
[itex]\frac{∂L}{∂y}[/itex]-[itex]\frac{d}{dx}[/itex] ([itex]\frac{∂L}{∂y'}[/itex])
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?