Need help seeing how they solved B1 and B2. Please show in detail.

2.44 Solve the following problem for x (1): x + 12x + 40x = 3 sin 51 x(0) = * (0) = 0 2. 44 7 + 12 x + 40 x 2 3 sinst x(o) = x(0) =0 roots = - 61 21 x (t ) = A, e sin2t + Aze oos 2t x (t ) = B, sinst + B, cosSt Substituting @ the diff. equation and *p( t ) = 5 B, cost - 58, sinst " Xact ) = - 25 B, sin ST - 25 B, cos 5t) solving for B, and By in the system gives : By = or By = - 4 Now X (t ), 1 = A, e sinat + Age cosat + 1 sin St - 4 cost 85 85 x (0 ) = 0 = A - 4 = ) A,= 4 x ( t ) = ( - 6 A , alt _ 2 A, est ) sin 2t + ( 2 A, @ - 6 Age ) cost + 5 cosSt + 20 sinst 85 * (0 ) = 0 = 2A, - 6 A, +5 ) A, = 19 170 X ( + ) = 19 - 6t e sin at + 4 -6t 170 85 - C cost + 1 sinst - 4 cos 5t 85 85