Question Number 102164 by mathmax by abdo last updated on 07/Jul/20

solve y^(′′) −y^′ +3y = e^(−x) sin(2x)

Answered by mathmax by abdo last updated on 08/Jul/20

(he)→y^(′′) −y^′ +3y =0→r^2 −r+3 =0 Δ =1−12 =−11 ⇒r_1 =((1+i(√(11)))/2) and r_2 =((1−i(√(11)))/2) ⇒y_h =a e^(r_1 x) +b e^(r_2 x) =e^(x/2) {acos(((√(11))/2)x) +b sin(((√(11))/2)x)} =au_1 +bu_2 W(u_1 ,u_2 ) = determinant ((( e^(x/2) cos((x/2)(√(11))) e^(x/2) sin((x/2)(√(11))))),(((1/2)e^(x/2) cos((x/2)(√(11)))−((√(11))/2)e^(x/2) sin(((x(√(11)))/2)) (1/2)e^(x/2) sin(((x(√(11)))/2))+((√(11))/2)e^(x/2) cos(((x(√(11)))/2))))) =(1/2)e^x {cos(((√(11))/2)x)sin(((√(11))/2)x) +(√(11))cos^2 (((√(11))/2)x)} −(1/2) e^x {sin(((√(11))/2)x)cos(((√(11))/2)x)−(√(11))sin^2 (((√(11))/2)x) =((√(11))/2)e^x w_1 = determinant (((o e^(x/2) sin(((√(11))/2)x) )),((e^(−x) sin(2x) .....)))=−e^(−(x/2)) sin(((√(11))/2)x)sin(2x) w_2 = determinant (((e^(x/2) cos(((√(11))/2)x) 0)),((..... e^(−x) sin(2x))))=e^(−(x/2)) cos(((√(11))/2)x)sin(2x) v_1 =∫ (w_1 /W)dx =−∫ ((e^(−(x/2)) sin(((√(11))/2)x)sin(2x))/(((√(11))/2) e^x )) =−(2/(√(11))) ∫ e^(−((3x)/2)) sin(((√(11))/2)x)sin(2x)dx (eazy to calculate) v_2 =∫ (w_2 /W)dx =∫ ((e^(−(x/2)) cos(((√(11))/2)x)sin(2x))/(((√(11))/2)e^x )) =(2/(√(11)))∫ e^(−((3x)/2)) cos(((√(11))/2)x)sin(2x)dx (eazy to calculate) so y_p =u_1 v_1 +u_2 v_2 and general soljtion is y =y_h +y_p