The question

5. Let X = {y E C'[0, In 2] : y(In 2) = 4} and consider the functional J : X -> R defined by In 2 2 dy Jly = + 272 dx da + (y(0) ) 2 + 43(In2) Find, from first principles, an extremum yo E X of J and show that yo is a minimum.5. The admissible variations are o are those C2 [0, In 2] functions satisfying 4 (In 2) = 0, AJ = = ( (uop' + 4yo4) dx + 2yo(0)4(0) ) += (242 + 242 ) dx + 4(0)2 ) , the natural boundary condition is -yo (0) + 2yo(0) = 0, the extremal is yo = e2z