First find solution of homogeneous problem.
r^(2)+2r=0 -> r_(1,2)=0, -2
Homogeneous solution:
y_(c)=c_(1)+c_(2)e^(-2t)
Let Y=At+B cos 2t+Csin 2t - because g(t)=3+4sin 2t
Plug Y into starting equation to find particular solution:
(At + B cos 2t + C sin 2t)''+2(At+Bcos 2t+Csin 2t)'=3+4sin 2t
-4Bcos 2t-4Csin 2t + 2A-4Bsin 2t+4Ccos 2t=3+4sin 2t
2A+cos 2t(-4B+4C)+sin 2t(-4C-4B)=3+4sin 2t
We obtain system:
2A=3 -> A=(3)/(2)
-4C-4B=4
-4B+4C=0 - add equations
-8B=4 -> B=-(1)/(2) -> C=-(1)/(2)
Y=(3)/(2)t-(1)/(2)cos 2t-(1)/(2)sin 2t
Solution of problem:
y=y_(c)+Y=c_(1)+c_(2)e^(-2t)+(3)/(2)t-(1)/(2)cos 2t-(1)/(2)sin 2t
Result:
y=c_(1)+c_(2)e^(-2t)+(3)/(2)t-(1)/(2)cos 2t-(1)/(2)sin 2t