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y''+a^2.y = sinx
The aux. equation
p^2+a^2 =0
p=ai or -ai
case 1: 0<a<1 or a>1
let
yg= Acos(ax) +Bsin(ax)
yp= Csinx +Dcosx
yp'= Ccosx -Dsinx
yp''=-Csinx -Dcosx
yp''+a^2.yp = sinx
-Csinx -Dcosx +a^2.(Csinx +Dcosx) = sinx
C(-1 +a^2)sinx +D(-1+a^2).cosx = sinx
C =1/(-1 +a^2) and D=0
通解
y=yg+yp = Acos(ax) +Bsin(ax) +[1/(-1 +a^2) ]sinx
case 2: a=1
p=i or -i
let
yg= Acosx +Bsinx
yp =x( Csinx +Dcosx )
yp' = Csinx +Dcosx + x( Ccosx -Dsinx )
yp''
=Ccosx -Dsinx + ( Ccosx -Dsinx ) +x( -Csinx -Dcosx )
=2Ccosx -2Dsinx +x( -Csinx -Dcosx )
yp''+yp = sinx
2Ccosx -2Dsinx +x( -Csinx -Dcosx ) +x( Csinx +Dcosx ) = sinx
2Ccosx -2Dsinx = sinx
D =-1/2 and C=0
y=yg+yp = Acosx +Bsinx -(1/2)xcosx
谢谢您的回复,好长的答案,我还得看一下,辛苦了!