Given that y1 (t) = cos(t) is a solution to y" - y' + y = sin(t) and y2 (t) = e 2t is a solution to y" - y + y = ed, use the superposition principle to find a particular solution to the differential equation y" - y' ty =3sin(t) - 15et. O yp (t) = cos(t) - 09 K -e 2t O yp (t) = 3 sin(t) - 15e2t O yp (t) = 3 cos(t) - 5e2t O yp (t) = = cos(t) - 3e2t