Example B The equation z(k+1, l) = 42(k, l+ 1) (5.84) leads to the conclusion that A = 4µ. Therefore, equation (5.84) has the special solutions 2(k, l) = 4*µk+e. (5.85) Summing expressions of this form allows the conclusion z(k, l) = 4* f(k + l), (5.86) where f is an arbitrary function of k + l. The separation-of-variables method, where zp(k, l) = CkDe, gives Ck+1De = 4CkDe+1; (5.87) which can be written in either of the forms Ck+1 De+1 Ck+1 or 4De+1 = a. (5.88) 4С = a De Ck De The respective solutions to these equations give z(k, l) = 4*ak+l or z(k, l) = 4-ak+e. (5.89) Summing over a allows us to obtain the general solution z(k, l) = 4* f (k + l) or z(k,l) = 4 g(k, l), (5.90) where f and g are arbitrary functions of k + l.

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