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Question Number 100344 by bemath last updated on 26/Jun/20
solve y′′+y = sin x
solvey+y=sinx
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
homogenoussolutionη2+1=0η=±iyh=Asinx+Bcosxparticularsolutionyp=x(Asinx+Bcosx)yp=Asinx+Bcosx+x(AcosxBsinx)yp=2Acosx2Bsinx+x(AsinxBcosx)A=0B=12yp=12xcosxGenerallsolutionyg=Asinx+Bcosx+12xcosx
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
y+y=sinx(he)r2+1=0r=+iyh=aeix+beix=αcosx+βsinx=αu1+βu2W(u1,u2)=|cosxsinxsinxcosx|=1W1=|0sinxsinxcosx|=sin2xW2=|cosx0sinxsinx|=cosxsinxv1=w1wdx=sin2x1dx=1cos(2x)2=x2+14sin(2x)v2=w2wdx=cosxsinxdx=12sin(2x)dx=14cos(2x)yp=u1v1+u2v2=cosx(x2+14sin(2x))+sinx(14cos(2x))=x2cosx+14cosxsin(2x)14sinxcos(2x)=x2cosx+12cos2xsinx14sinx(2cos2x1)=x2cosx+14sinxthegeneralsolutionisy=yh+yp=αcosx+βsinxx2cosx+14sinx

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