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 !!)