If a third order, linear, homogeneous, constant coefficients differential equation has e²ᵗ+sin(t) as a solution, then the solution of the differential equation that satisfies y(0)=1,y′(0)=2,y′′(0)=0, is given by

A. y(t)=1/5 e²ᵗ+4/5​cos(t)+58​sin(t)
B. y(t)=4/5​ e²ᵗ+1/5 cos(t)+52​sin(t)
C. y(t)=8/5​ e²ᵗ−3/5 ​cos(t)−56​sin(t)
D. y(t)=1/2 ²ᵗ+ 1/2​cos(t)+sin(t)
E. y(t)= 3/4 e²ᵗ+1/4​cos(t)+21​sin(t)