Question

例设f"(x)+f(x)=-xsinx+2cosx其中f(x)

Answer 1

f''(x)+f(x)=-xsinx+2cosx
The aux. equation
r^2 +1 =0
r=i or -i
let
fg(x) = Acosx +Bsinx
fp(x)= x^2.( Ccosx +Dsinx) + x.(Fcosx+Gsinx)
fp'(x)
= x^2.( -Csinx +Dcosx) + 2x.( Ccosx +Dsinx) + x.(-Fsinx+Gcosx) + (Fcosx+Gsinx)
=x^2.( -Csinx +Dcosx) + x.[ (2C+G)cosx +(2D-F)sinx)] + (Fcosx+Gsinx)
fp''(x)
=x^2.( -Ccosx -Dsinx)+2x.( -Csinx +Dcosx)
+ x.[ -(2C+G)sinx +(2D-F)cosx)] +[ (2C+G)cosx +(2D-F)sinx)] + (-Fsinx+Gcosx)
=x^2.( -Ccosx -Dsinx) + x[ (4D-F)cosx +(-4C-G)sinx] +[ (2C+2G)cosx +(2D-2F)sinx)]
f''p(x)+fp(x)=-xsinx+2cosx
x^2.( -Ccosx -Dsinx) + x[ (4D-F)cosx +(-4C-G)sinx] +[ (2C+2G)cosx +(2D-2F)sinx)]
+ x^2.( Ccosx +Dsinx) + x.(Fcosx+Gsinx) =-xsinx+2cosx
x[ 4Dcosx -4Csinx] +[ (2C+2G)cosx +(2D-2F)sinx)] =-xsinx+2cosx
coef. of xcosx => D=0
coef. of xsinx => C=1/4
coef. of cosx :
2C+2G =2
1+2G =2
G=1/2
coef. of sinx :
2D-2F =0
F=0
ie
fp(x)
= x^2.( Ccosx +Dsinx) + x.(Fcosx+Gsinx)
= (1/4)x^2.cosx + (1/2)xsinx
通解
f(x) = fg(x) +fp(x) = Acosx +Bsinx + (1/4)x^2.cosx + (1/2)xsinx