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A particular solution to the differential equation:

$$\begin{equation} y'' + 4y = \sin 2t \end{equation}$$

...is:

$y(t) = -\cfrac{t}{4} \cos 2t$

$y(t) = -\cfrac{1}{4} \cos 2t$

$y(t) = -\cfrac{t}{4} \sin 2 t$

$y(t) = -\cfrac{1}{4} \sin 2t$