D2+DD'-2D'2=(y-1)ex
D=m
D'=1
m2+m-2=0
m2-m+2m-2=0
(m+2)(m-1)=0
m=-2 or 1
The roots remain distinct
C.F=f1(y-2x)+f2(y+x)
P.I=D2+DD′−2D′21•(y−1)ex
=(D+2D′)(D−D′)1•(y−1)ex
=[D+2D′1][D−D′1•(y−1)ex]
D-mD'=D-D'
∴m=1
y=C-mx=C-x
P.I=D+2D′1[∫(C−x−1)exdx]
=D+2D′1[(C−x−1)ex−ex]
=D+2D′1[(y−1)ex−ex]
D-mD'=D+2D'
∴m=−2
y=C-mx=C+2x
P.I=∫((C+2x−1)ex−ex)dx
=(C+2x-1)ex-(2)ex--ex
=(y-1)ex--2ex--ex
=(y-4)ex
Complete solution,Z=C.F+P.I
=f1(y−2x)+f2(y+x)+(y−4)ex
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