The solution to the Initial value problem, x''+2x'+5x=0, is the sum of the steady periodic solution x_sp and x_tr. Find both.
The Attempt at a Solution
I already found x_sp ( the particular solution). It is (-44/533)cos(7t)+(14/533)sin(7t).
x_tr = e^(-t)(Acos(2t)+Bcos(2t))
x(t) = e^(-t)(Acos(2t)+Bsin(2t)) + (-44/533)cos(7t)+(14/533)sin(7t).
x'(t) = e^(-t)(-2Asin(2t)+2Bcos(2t)) - e^(-t)(Acos(2t)+Bsin(2t)) +
x(0)=1(A+0)-(44/553) ==> A = 44/533
x'(0)= 1(2B)-1(A)+(98/533) ==> B= -27/533
this is not right however. Can someone see where I might have messed up?