Solve y'' +2y' + 5y = cos(2t).

I found the Yc = e^-t(A cos(2t) + Bsin(2t)).

I am finding trouble in working out the particluar integral. This is what i have done so far:

Yp= t(Acos(2t) + Bsin(2t))

Y'p= t(-2Asin(2t) + 2Bcos(2t))+ Acos(2t) + Bsin(2t)

Y''p = t(-4Acos(2t) - 4Bsin(2t) + 2(-2Asin(2t) + 2Bcos(2t))

Sub this back into equation:

y'' +2y' + 5y=

t(-4Acos(2t) - 4Bsin(2t) + 2(-2Asin(2t) + 2Bcos(2t)) + 2(t(-2Asin(2t) + 2Bcos(2t))+ Acos(2t) + Bsin(2t)) + 5(t(Acos(2t) + Bsin(2t)).

Now, I dont know how this should be simplified because of the 't' function and how I can equate this to find out A and B.

Thanks