Question Number 100344 by bemath last updated on 26/Jun/20

solve y′′+y = sin x

Answered by bobhans last updated on 26/Jun/20

homogenous solution  η^2 +1 = 0 ⇒η =± i   y_h  = Asin x + Bcos x  particular solution   y_p  = x(Asin x+ Bcos x)   y_p ′= Asin x+Bcos x+x(Acos x−Bsin x)  y_p ′′= 2Acos x−2Bsin x+x(−Asin x−Bcos x)  A=0 ∧ B=(1/2)  y_p  = (1/2)xcos x   Generall solution   y_g  = Asin x+Bcos x+(1/2)x cos x

Answered by mathmax by abdo last updated on 26/Jun/20

y^(′′)  +y =sinx   (he) →r^2 +1 =0 ⇒r =+^− i ⇒y_h =ae^(ix)  +be^(−ix)  =αcosx +βsinx  =αu_1  +β u_2   W(u_1  ,u_2 ) = determinant (((cosx           sinx)),((−sinx       cosx)))=1  W_1 = determinant (((0           sinx)),((sinx    cosx)))=−sin^2 x  W_2 = determinant (((cosx        0)),((−sinx    sinx)))=cosx sinx  v_1 =∫ (w_1 /w)dx =−∫ ((sin^2 x)/1)dx =−∫((1−cos(2x))/2) =−(x/2)+(1/4)sin(2x)  v_2 =∫ (w_2 /w)dx =∫ cosx sinx dx =(1/2)∫ sin(2x)dx =−(1/4)cos(2x) ⇒  y_p =u_1 v_1  +u_2 v_2 =cosx(−(x/2)+(1/4)sin(2x))+sinx(−(1/4)cos(2x))  =−(x/2)cosx +(1/4)cosx sin(2x)−(1/4)sinx cos(2x)  =−(x/2)cosx +(1/2)cos^2 x sinx −(1/4)sinx(2cos^2 x−1)  =−(x/2)cosx +(1/4)sinx  the general solution is  y =y_h  +y_p =αcosx +β sinx −(x/2)cosx +(1/4)sinx