Question

高等数学求特解

Answer 1

设y=(ax+b)cos2x+(cx+d)sin2x是y''+y=xcos2x①的解，则
y'=acos2x-2(ax+b)sin2x+csin2x+2(cx+d)cos2x
=(2cx+a+2d)cos2x+(-2ax-2b+c)sin2x,
y''=2ccos2x-2(2cx+a+2d)sin2x-2asin2x+2(-2ax-b+c)cos2x
=(-4ax-2b+4c)cos2x+(-4cx-4a-4d)sin2x,
都代入①，得(-3ax-b+4c)cos2x+(-3cx-4a-3d)sin2x=xcos2x,
比较得-3ax-b+4c≡x,且-3cx-4a-3d≡0,
∴a=-1/3,b=c=0,d=4/9.
∴所求特解是y=(-x/3)cos2x+(4/9)sin2x.