
Yefim S. answered 10/14/23
Math Tutor with Experience
4r2 + 16.8 = 0; r2 = - 4.2; r = ±√4.2i; xgh = Ccos√4.2t + Dsin√4.2t
xp = Acos2t + Bsin2t; xp' = - 2Asin2t + 2Bcos2t; xp'' = - 4Acos2t - 4Bsin2t;
4Acos2t - 4Bsin2t + 16.8Acos2t + 16.8Bsin2t = cos2t; 12.8Acos2t + 12.8Bsin2t = cos2t;
B = 0; A = 1/12.8 = 5/64. So, xp = 5/64cos2t;
x = 5/64cos2t + Ccos√4.2t + Dsin√4.2t.
Because √4.2 ≈ 2. We have x ≈ (5/64 + C)cos2t Dsin2t and frequency f ≈ 1/π