Given that y1(t)=cos(t) is a solution to y''-y'y=sin(t) and y2(t)=1/3e²t is a solution to y''-y'y=e²t. What is the general solution to the differential equation y''-y'y=sin(t)+e²t?

A) y(t) = C1 cos(t) + C2 e²t + sin(t)
B) y(t) = C1 sin(t) + C2 e²t + cos(t)
C) y(t) = C1 cos(t) + C2 e²t - sin(t)
D) y(t) = C1 sin(t) + C2 e²t - cos(t)