Question

微分方程通解问题

Answer 1

y''-2y'+y = (1/2)(1-cos2x)
The aux. equation
p^2-2p+1=0
p=1
let
yg= (A+Bx)e^x
yp= Csin2x+ Dcos2x +E
yp' =2Ccos2x- 2Dsin2x
yp'' =-4Csin2x- 4Dcos2x
yp''-2yp'+yp = (1/2)(1-cos2x)
-4Csin2x- 4Dcos2x -2[2Ccos2x- 2Dsin2x] +Csin2x+ Dcos2x +E =(1/2)(1-cos2x)
(-3C+4D)sin2x + (-4C-3D)cos2x + E =(1/2)(1-cos2x)
=> E = 1/2
and
-3C+4D = 0 (1)
-4C-3D = -1/2 (2)

3(1) +4(2)
-25C =-2
C=2/25
from (1)
-3C+4D = 0
-6/25 +4D=0
D= 3/50
ie
yp= (2/25)sin2x+ (3/50)cos2x +1/2
y
= yg+ yp
= (A+Bx)e^x +(2/25)sin2x+ (3/50)cos2x +1/2

Answer 2