Hi, please see the attached question which I am having trouble with and need help with.
I have done part a) and part b)
the E-L equation for part a) I got to be 2y''sinhx - 2y'coshx + ysinh^3(x) = 0
and for part b) the new variable I have is u = cosh x and with new limits u1 = cosh a and
u2 = cosh b
and for part c) I've managed to get as far as the E-L equation being 2y''- y = 0 with a general solution of
y(u)=Acos(u/sqrt2) + Bsin(u/sqrt2)
which in terms of x is y(x) = Acos(coshx/sqrt2) + Bsin(coshx/sqrt2)
I would really appreciate some help on the rest of part c) and part d) and if you could also check that what I have for parts a), b) and my partial part c) is correct.
Thank you.© BrainMass Inc. brainmass.com October 10, 2019, 8:16 am ad1c9bdddf
(a) The associated Euler-Lagrange equation is:
By differentiating, we get:
The equation Euler-Lagrange becomes:
or, by multiplying with - sinh2(x):
(b) Let's consider a variable change of form y = y(u) where
The given functional changes as follows:
A problem of calculus of variations is solved using a variable change which simplifies the differential equation obtained by applying Euler-Lagrange equation for the given functional which has to be minimized (or maximized).