Quote by BomboshMan
and that the boundary conditions for antinodes at x = 0 and x = L are
D(0,t) = ±2a
D(L,t) = ±2a

That looks wrong. I think it should be
##\partial D(0,t)/\partial x## = 0
##\partial D(1,t)/\partial x## = 0
which gives ##D(0,t) = D(1,t) = 2a \cos(\omega t)##.