LULU ALIIS SUILLIUI. Solve the variational problem with one variable end for the functional JW = / 2xy + (%)' da; y(0) = 0. 0 Note that the boundary condition is specified only at one end, at x = 0. At the other end, = 2, the function may have any value, hence the name "the problem with one variable end" As we demonstrated in class, one has to follow an algorithm similar to that we applied to problems with two fixed ends: • Write down the Euler-Lagrange equation. . At the variable end, require that the solution y(a) of the Euler-Lagrange equation satisfies the boundary condition = 0 Oy In this problem, b = 2. Using that condition and the boundary condition at the other end r = 0, solve the Euler-Lagrange equation. • Analyse if the solution y(2) gives an extremum and, if yes, the type of the extre- mum, by considering the difference Jy() + n()] - Jly()]. Hint for solving the differential equation: to solve a linear differential equation with a right-hand-side part you may guess one of its solutions and add it to the generic solution of the respective linear differential equation without the right-hand side.