Final answer:
The ODE boundary value problem y" + 4y = 0, with the condition y(L) = 1, fails to have a solution for any L if the proper boundary conditions for a particle in a box are y(0) = y(L) = 0, due to a typo in the question's conditions.
Step-by-step explanation:
The question asks about the existence of solutions for a boundary value problem described by the ordinary differential equation (ODE) y" + 4y = 0, with boundary conditions y(0) = 0 and y(L) = 1, where L > 0. According to the provided information, the particle described by this equation cannot exist beyond the walls, which implies that y(0) = y(L) = 0. However, there is a discrepancy in the boundary conditions given in the question (y(L) = 1) and the conditions that would logically apply to a particle in a box (y(L) = 0). If we stick to the logical conditions implied by the physics context, the solution to the ODE only exists if the boundary conditions are y(0) = 0 and y(L) = 0. For the ODE given, the general solution is of the form A sin(2x) + B cos(2x). Applying the initial conditions, A = 0 and B = 0, to comply with the boundary conditions at x = 0 and x = L. Without a typo, this would lead to the trivial solution y(x) = 0 only, thus failing to meet the condition y(L) = 1 for any L, indicating no nontrivial solution exists. If the boundary condition y(L) = 1 is correct, then the problem would have solutions for certain values of L, specifically those that align with the resonance conditions of the sinusoidal solutions within the interval [0, L]. But if we must have y(L) = 0 as per the context-provided conditions, the problem fails to have a solution for any L based on the typo in boundary conditions.