The given equation can be written as
(D3–D′3)z=x3y3 --------(1)
Corresponding auxiliary equation is
m3–1=0 , i.e (m–1)(m2+m+1)=0
or
(m–1)(m–w)(m–w2)=0, for w to be the cube root of unity----(2)
C.F.=φ1(y+x)+φ2(y+wx)+φ3(y+w2x) -----(3)
Now
p.I=D3−D′31x3y3=D3[1−(DD′)3]1x3y3
=D31[1−(DD′)3]−1x3y3=D31[1+DD′3+...]x3y3
=D31x3y3+D6D′3x3y3
(on taking DD′ to the powers to which y is appearing in
F(x,y)=xnym i.e neglecting terms containing powers more than 3 in D′
=4⋅5⋅6x6y3+4⋅5⋅6⋅7⋅8⋅96x9
P.I=120x6y3+10080x9
Hence complete solution
z=φ1(y+x)+φ2(y+wx)+φ3(y+w2x)+120x6y3+10080x9
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