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