Question

(7)微分方程 y''+4y=2cos^2x 的特解形式为

Answer 1

y''+4y=2(cosx)^2
y''+4y=cos2x +1
The aux. equation
r^2+4=0
r=2i or -2i
yg=Acos2x+Bsin2x
yp=x(Ccos2x+Dsin2x) +F
yp''=
y''+4y=2(cosx)^2
y''+4y=cos2x +1
The aux. equation
r^2+4=0
r=2i or -2i
yg=Acos2x+Bsin2x
yp=x(Ccos2x+Dsin2x) +E
yp'=(Ccos2x+Dsin2x) +x(-2Csin2x +2Dcos2x)
yp''
=(-2Csin2x+2Dcos2x) +(-2Csin2x +2Dcos2x) +x(-4Ccos2x -4Dsin2x)
=(-4Csin2x+4Dcos2x)+x(-4Ccos2x -4Dsin2x)
yp''+4yp=cos2x +1
(-4Csin2x+4Dcos2x)+x(-4Ccos2x -4Dsin2x) +4[x(Ccos2x+Dsin2x) +E]=cos2x +1
(-4Csin2x+4Dcos2x)+4E=cos2x +1
=>
C=0, D=1/4, E=1/4
yp=x(Ccos2x+Dsin2x) +F = (1/4)xsin2x +1/4
通解
y=yg+yp=Acos2x+Bsin2x +(1/4)xsin2x +1/4