Question Number 104651 by Ar Brandon last updated on 23/Jul/20

Consider the differential equation;  x′′(t)+x(t)=t^2 cos(2t).  Knowing that the Complementary Function is;  y_(CF) =acos(t)+bsin(t),  In what form can the particular integral be expressed  so as to obtain the general solution to the given   differential equation?

Answered by mathmax by abdo last updated on 23/Jul/20

y_h =acost +bsint  =au_1 (t)+bu_2 (t)  W(u_1  ,u_2 ) = determinant (((cost         sint)),((−sint      cost)))=cos^2 t +sin^2 t =1 ≠0  W_1 = determinant (((0         sint)),((t^2 cos2t  cost)))=−t^2  sint cos(2t)  W_2 = determinant (((cost           0)),((−sint      t^2 cos(2t))))=t^2  cost cos(2t)  v_1 =∫ (w_1 /w)dx =−∫ t^2  sint cos(2t)dt  we have sint cos(2t) =cos((π/2)−t)cos(2t)  =(1/2) { cos((π/2)−t +2t) +cos((π/2)−t−2t)}=(1/2){−sint +sin(3t)} ⇒  v_1 =−(1/2)∫(−sint +sin(3t)t^2  dt =(1/2) ∫ t^2  sint dt−(1/2) ∫t^2 sin(3t)dt  (1/2)∫t^2 sint dt=(1/2){  −t^(2 ) cost +∫ 2t cost dt} =−(t^2 /2)cost +2 {  tsint −∫ sint dt}  =−(t^2 /2)cost +2{ tsint +cost}  ∫t^2  sin(3t)dt =_(3t =u)   ∫(u^2 /9)sin(u)(du/3) =(1/(27)) ∫ u^2  sinu du =...  v_2 =∫ ((w2)/w)dx =∫ t^2  cost cos(2t)dt =(1/2)∫ t^2 {cos(3t) +cos(t)}dt=...  the particular solution is x_p =u_1 v_1  +u_2  v_2  and the general solution is  x (t) =x_h (t)+x_p (t)

Commented byAr Brandon last updated on 23/Jul/20

Thanks for your time Sir. But actually I needed  to know how to express the Particular Integral.  Can it be;  y_(PI) =(a_1 t^2 +b_1 t+c_1 )cos(2t)+(a_2 t^2 +b_2 t+c_2 )sin(2t) ?