Solution of $$(D^2 + 16)y = \cos 4x,$$ is

A

$$y = A \cos 4x + B \sin 4x + \frac{1}{8} \cos 4x$$

B

$$y = A \cos 4x + B \sin 4x + \frac{x}{8} \sin 4x$$

C

$$y = A \cos 4x + B \sin 4x + \frac{1}{8} \sin 4x$$

D

$$y = A \cos 4x + B \sin 4x + \frac{x}{8} \cos 4x$$