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Solution of $$(D^2 + 16)y = \cos 4x,$$ is
$$y = A \cos 4x + B \sin 4x + \frac{1}{8} \cos 4x$$
$$y = A \cos 4x + B \sin 4x + \frac{x}{8} \sin 4x$$
$$y = A \cos 4x + B \sin 4x + \frac{1}{8} \sin 4x$$
$$y = A \cos 4x + B \sin 4x + \frac{x}{8} \cos 4x$$
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