Question

求y''+y=sinx+xcos2x的通解，急求详细过程

Answer 1

y''+y=sinx+xcos2x
y''+y=0的特征根：±i
对于方程y''+y=sinx,特解形式y=x(Asinx+Bcosx)
对于方程y''+y=xcos2x,特解形式y=(Cx+D)(Esin2x+Fcos2x)
故原方程特解形式y=x(Asinx+Bcosx)+(Cx+D)(Esin2x+Fcos2x)
代入求出ABCDEF
通y=C1sinx+C2cosx+x(Asinx+Bcosx)+(Cx+D)(Esin2x+Fcos2x)

追问

可以再把ABCDE那些求一下吗，这些在作业帮我也搜到了

追答

y=C1cosx+C2sinx-1/2xcosx-1/3xcos2x+4/9sin2x

Answer 2

y''+y=sinx+xcos2x
The aux. equation
p^2 +1 =0
p=i or -i
let
yg= Acosx +Bsinx
let
yp = (Cx+D)sinx + (Ex+F)cos2x
yp'
= (Cx+D)cosx + Csinx - 2(Ex+F)sin2x + Ecos2x
yp''
=-(Cx+D)sinx + Ccosx + Ccosx - 4(Ex+F)cos2x -2Esin2x
=-(Cx+D)sinx + 2Ccosx - 4(Ex+F)cos2x -2Esin2x
yp''+yp=sinx+xcos2x
[-(Cx+D)sinx + 2Ccosx - 4(Ex+F)cos2x -2Esin2x] +[(Cx+D)sinx + (Ex+F)cos2x]
=sinx+xcos2x
-3(Ex+F)cos2x -2Esin2x+ 2Ccosx =sinx+xcos2x
coef. of cosx => C=0
coef. of sin2x => E=0
coef. of xcos2x =>E =-1/3
coef. of cos2x => F=0
yp = (Cx+D)sinx + (Ex+F)cos2x = Dsinx -(1/3)xcos2x
通解
y
=yg+yp
= Acosx +Bsinx +Dsinx -(1/3)xcos2x
=Acosx + Fsinx -(1/3)xcos2x

追问

朋友，为啥有的我看不到，那个coef啥的

看不懂

追答

coef. = coefficient =系数
coef. of cosx : cosx 的系数