设特解y=asinx+bcosx
y'=acosx-bsinx
y'-y=(acosx-bsinx)-(asinx+bcosx)
=(-a-b)sinx+(a-b)cosx=cosx
比较对应项系数,得-a-b=0,a-b=1
解得a=1/2,b=-1/2
所以特解y=(1/2)*sinx-(1/2)*cosx