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

In summary, the Euler-Lagrange equation for the given integral is 2y*sin(x) - y'^2 - 2yy'. The discrepancy between the hand computation and the Mathematica solution is likely due to different input or calculation methods.
  • #1
~Sam~
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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?
 
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  • #2
~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?

I get ##L_y - D_x L_{y'} = 2 y \sin(x) - y'^2 - 2 y y''##.
 
  • #3
Ray Vickson said:
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
 
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  • #4
~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

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.
 
  • #5
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.

Regardless, I assume that's 2ysin(x)−y′2−2yy′ correct Euler-Lagrange then?
 
  • #6
~Sam~ said:
Regardless, I assume that's 2ysin(x)−y′2−2yy′ correct Euler-Lagrange then?

Yes it is.
 

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

What are Euler-Lagrange equations?

Euler-Lagrange equations are a set of differential equations used to find the stationary points of a functional. They are commonly used in the fields of mathematics, physics, and engineering.

What is the significance of Euler-Lagrange equations?

Euler-Lagrange equations are significant because they allow us to find the minimum or maximum value of a functional. This is useful in many applications, such as finding the shortest path between two points or determining the path of a particle that minimizes energy.

How do Euler-Lagrange equations work?

Euler-Lagrange equations are derived from the calculus of variations, which is a branch of mathematics that deals with finding the extremum of a functional. The equations involve taking derivatives of the functional with respect to the variables and setting them equal to zero.

What are some real-world applications of Euler-Lagrange equations?

Euler-Lagrange equations have a wide range of applications in physics, engineering, and economics. They are used in the study of mechanics, electromagnetism, fluid dynamics, and control theory, among others.

Are there any limitations to Euler-Lagrange equations?

Euler-Lagrange equations have limitations in cases where the functional is not differentiable or the boundary conditions are not well-defined. They also do not provide a global minimum or maximum, only local extrema. In some cases, other methods such as the Hamiltonian formalism may be more suitable.

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