I think the question was
$$
y'''-y= 2 \sin t
$$
solution:
$$
r^3 - 1 = 0 \\
(r+1)(r^2+r+1) = 0\\
r = -1, \frac{-1 \pm \sqrt{3} i}{2}\\
y_h = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} \\
y_p = A \sin t + B \cos t \\
y_p' = A \cos t - B \sin t \\
y_p'' = -A \sin t - B \cos t \\
y_p''' = - A \cos t + B \sin t \\
A = -1, B = 1\\
y = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} + \cos t - \sin t
$$
