Given that the Euler-Lagrange equation (not the usual physics form) is
dxd(∂y′∂F)−∂y∂F=0
show that if F does not explicitly depend on x, then the Euler equation can be integrated as
F−y′∂y′∂F=C.
Solution
Part 1.
We must first show that dxd(F−y′∂y′∂F)−∂x∂F=0.
We observe that dxdF=∂x∂F+∂y∂Fdxdy+∂y′∂Fdxdy′=∂x∂F+∂y∂Fy′+∂y′∂Fy′′.
Since dxd(y′∂y′∂F)=y′dxd(∂y′∂F)+∂y′∂Fy′′.
By subtraction dxd(F−y′∂y′∂F)=∂x∂F+y′[∂y∂F−dxd(∂y′∂F)].
Since dxd(∂y′∂F)−∂y∂F=0, we find that
dxd(F−y′∂y′∂F)−∂x∂F=0.
Part 2.
Since F does not explicitly depend on x, then ∂x∂F=0.
By integration
F−y′∂y′∂F=C for some arbitrary constant C.
Check out my other notes at Proof, Disproof, and Derivation
#Calculus
#CalculusofVariation
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