Question

求微分方程的特解

题目如图：

Answer 1

y''+y=xcos2x
y''+y=0
特征方程
r^2+1=0
r=i或r=-i
y=Ccosx+C1sinx

设方程特解
y=(mx+n)cos2x +(sx+t)sin2x
y'=mcos2x-(2mx+2n)sin2x +ssin2x +(2sx+2t)cos2x
y''= -2msin2x-2msin2x-(4mx+4n)cos2x +2scos2x+2scos2x-(4sx+4t)sin2x
=(-4m-4t)sin2x+(2s-4n)cos2x -4mxcos2x-4sxsin2x

(-4m-3t)sin2x+(2s-3n)cos2x-3mxcos2x-3sxsin2x=xcos2x
-3m=1 -3s=0
m=-1/3 s=0
t=4/9 n=0
特解y=(-1/3)xcos2x+(4/9)sin2x
y'=(5/9)cos2xx+(2/3)xsin2x
y''=(-10/9)sin2x+(4/3)xcos2x+(2/3)sin2x
y''+y=xcosx

通解y=(-1/3)xcos2x+(4/9)sin2x+Ccosx+C1sinx

Answer 2