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