Question Number 104034 by mohammad17 last updated on 19/Jul/20

Solve: y^(′′) +2y^′ +2y=secax

Answered by bramlex last updated on 19/Jul/20

HE : ℓ^2 +2ℓ+2=0  (ℓ+1)^2 +1=0 →ℓ=−1±i   y_h  = e^(−x) (C_1 cos x+C_2 sin x)

Commented bymohammad17 last updated on 19/Jul/20

sir i want yp ?

Commented bybobhans last updated on 19/Jul/20

y_1 = e^(−x) cos x →y_1 ′=−e^(−x) cos x−e^(−x) sin x                         →y_1 ′=−e^(−x) (cos x+sin x)  y_2 = e^(−x) sin x→y_2 ′=−e^(−x) sin x+e^(−x) cos x                          →y_2 ′=−e^(−x) (sin x−cos x)  W(y_1 ,y_2 )= determinant (((            e^(−x) cos x                                  e^(−x) sin x)),((−e^(−x) (cos x+sin x)      −e^(−x) (sin x−cos x)))  = −e^(−2x) (cos xsin x−cos ^2 x)+e^(−2x) (sin x  cos x +sin ^2 x) = e^(−2x)   I_1 = e^(2x) ∫ e^(−x) cos x.sec x dx = −e^x   I_2 = e^(2x)  ∫ e^(−x) sin x.sec x dx   I_2 =(1/2)e^(2x)  ∫ e^(−x) sin 2x dx   y_p  = −y_1 I_2 + y_2 I_1

Answered by mathmax by abdo last updated on 19/Jul/20

y^(′′)  +2y^′  +2y =(1/(cos(x)))  h→r^2  +2r +2 =0→Δ^′  =−1 ⇒r_1 =−1+i and r_2 =−1−i ⇒  y_h =ae^((−1+i)x)  +b e^((−1−i)x)  =e^(−x) {αcosx +βsinx} =αu_1  +βu_2   W(u_1  ,u_2 ) = determinant (((e^(−x)  cosx          e^(−x) sinx)),((−(cosx+sinx)e^(−x)     (cosx −sinx)e^(−x) )))  =e^(−2x) {cos^2 x−cosx sinx}+e^(−2x) {sin^2 x+sinx cosx} =e^(−2x)  ≠0  W_1 = determinant (((o                e^(−x) sinx)),(((1/(cos(ax)))        (cosx−sinx)e^(−x) )))=−e^(−x)  ((sinx)/(cos(x)))  W_2 = determinant (((e^(−x) cosx                                  0)),((−(cosx +sinx)e^(−x)        (1/(cosx)))))=e^(−x)    v_1 =∫ (w_1 /w)dx =−∫  ((e^(−x)  sinx)/(e^(−2x)  cosx))dx =−∫ e^x  ((sinx)/(cosx))dx  v_2 =∫ (w_2 /w)dx =∫  (e^(−x) /e^(−2x) )dx =∫ e^x  dx =e^x   ⇒  y_p =u_1 v_1  +u_2 v_2 =−e^(−x)  cosx ∫ ((e^x sinx)/(cosx))dx + sinx   the general solution is y =y_h  +y_p