Equation (5.119) rewritten with the assumed form z(k, l) = CkDe is Ck+3 De – 3Ck+2De+1+3Ck+1De+2 – CkDe+3 = 0. (8.30) Dividing all terms by Ck Dk gives ((똥)- (똥) () Ck+3 Ck+2 Ck+1 +3 De+2 De De+1 De+3 De = 0. (8.31) We can start with either one of the following choices Ck+1 De+1 or (8.32) = a. Ck De Using the second expression gives De = Aa', A = arbitrary function of a. (8.33) %3D Substitution of this result in equation (8.31) yields Ck+3 Ck+2 3a + 3a? Ck (Ot1) - a' = 0, Ck+1 (8.34) Co or Ck+3 – (3a)Ck+2 + 3a²Ck+1 – a³Ch = 0. (8.35) The corresponding characteristic equation is p3 – (3a)r2 + (3a²)r – a³ = (r – a)³ = 0. (8.36) | Therefore, C(a) is Ck(a) = A1(a)a* + A2(a)ka* + A3(@)k²a* (8.37) and z(k, l, a) = Cr(@)D¿(a) %3D = á(a)a*+l + Ã2(a)ka*+l + Ã3(a)k²a*+e. (8.38) If we sum/integrate over a, we obtain the following solution to equation (5.119) 2(k, e) = f(k + l) + kg(k + l) + k²h(k +l), (8.39) where (f, g, h) are arbitrary functions of (k + e).

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