Question Number 100017 by bemath last updated on 24/Jun/20

(d^2 y/dx^2 ) + y = sec 3x

Answered by mathmax by abdo last updated on 24/Jun/20

y^(′′)  +y =(1/(cos(3x)))     (he) →r^2  +1 =0 ⇒r =+^− i ⇒  y_h =ae^(ix)  +b e^(−ix)  =α cosx +βsinx =αu_(1 ) +βu_2   W(u_1  ,u_2 ) = determinant (((cosx          sinx)),((−sinx        cosx)))=1  W_1 = determinant (((0            sinx)),(((1/(cos(3x) ))   cosx)))=−((sinx)/(cos(3x)))  W_2 = determinant (((cosx        0)),((−sinx    (1/(cos(3x))))))=((cosx)/(cos3x))  v_1 =∫ (w_1 /w)dx =−∫ ((sinx)/(cos(3x)))dx  v_2 =∫ (w_2 /w)dx =∫((cosx)/(cos(3x)))dx ⇒y_p =u_1 v_1  +u_2 v_2   and y =y_h  +y_p     let try ∫  ((sinx)/(cos(3x)))dx we have  cos(3x) =cos(2x+x) =cos(2x)cosx −sin(2x)sinx  =(2cos^2 x−1)cosx −2sin^2 x cosx =2cos^3 x−cosx −2(1−cos^2 x)cosx  =2cos^3 x−cosx−2cosx +2cos^3 x =4cos^3 x−3cosx ⇒  ∫  ((sinx)/(cos(3x)))dx =∫ ((sinxdx)/(4cos^3 −3cosx)) =_(cosx=t)  −∫  (dt/(4t^3 −3t))  =−∫  (dt/(t(4t^2 −3))) =−(1/4)∫  (dt/(t(t−((√3)/2))(t+((√3)/2)))) and  f(t) =(1/(t(t−((√3)/2))(t+((√3)/2)))) =(a/t) +(b/(t−((√3)/2))) +(c/(t+((√3)/2)))  (eazy to find a ,b and c) ⇒  ∫ ((sinx)/(cos(3x)))dx =−(1/4){ aln∣t∣+bln∣t−((√3)/2)∣ +cln∣t+((√3)/2)∣}  =−(1/4){aln∣cosx∣+bln∣cosx−((√3)/2)∣ +c ln∣t+((√3)/2)∣} +c(dont add c in y_p !!)