insynC
Jan21-09, 05:57 PM
Just did this in class today and was doing a problem to see if I understood it and I'm not sure I did. Thanks for any help
1. The problem statement, all variables and given/known data
Solve the Euler equation to make the following integral:
(integral from x1->x2)
∫ [(y')² + y²] dx
2. Relevant equations
Euler-Lagrange equation
∂F/∂y - d/dx (∂F/∂y') = 0
3. The attempt at a solution
Clearly F = (y')² + y²
In class we had been rearranging the integral so that ∂F/∂y = 0, which made the problem much simpler. I was unable to do this here as there was no apparent way to introduce ds to swap y' for x'. This was my concern as, although I didn't think all equations would be this simple, I thought most would reduce.
So instead I proceeded as follows:
∂F/∂y - d/dx (∂F/∂y') = 2y - d/dx (2y') = 2y - 2y'' = 0 => y'' = y
And this is easy enough to solve. But I'm concerned I've made a mistake getting there.
Have I made a mistake up to here or is there a better way to proceed?
Thanks
1. The problem statement, all variables and given/known data
Solve the Euler equation to make the following integral:
(integral from x1->x2)
∫ [(y')² + y²] dx
2. Relevant equations
Euler-Lagrange equation
∂F/∂y - d/dx (∂F/∂y') = 0
3. The attempt at a solution
Clearly F = (y')² + y²
In class we had been rearranging the integral so that ∂F/∂y = 0, which made the problem much simpler. I was unable to do this here as there was no apparent way to introduce ds to swap y' for x'. This was my concern as, although I didn't think all equations would be this simple, I thought most would reduce.
So instead I proceeded as follows:
∂F/∂y - d/dx (∂F/∂y') = 2y - d/dx (2y') = 2y - 2y'' = 0 => y'' = y
And this is easy enough to solve. But I'm concerned I've made a mistake getting there.
Have I made a mistake up to here or is there a better way to proceed?
Thanks