What is the Correct Euler-Lagrange for ∫y(y')2+y2sinx dx?

[~Sam~]

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]

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?

I get $L_y - D_x L_{y'} = 2 y \sin(x) - y'^2 - 2 y y''$.

 

[~Sam~]

I get $L_y - D_x L_{y'} = 2 y \sin(x) - y'^2 - 2 y y''$.

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]

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

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~]

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.

Regardless, I assume that's 2ysin(x)−y′²−2yy′ correct Euler-Lagrange then?

 

[Dick]

Regardless, I assume that's 2ysin(x)−y′²−2yy′ correct Euler-Lagrange then?

Yes it is.