5.5.2 Еxample B Consider the equation z(k + 1, l + 1) + 2z(k, e) = k2 + l +1, (5.170) or (E, E2 + 2)z(k, l) = k² + l+ 1. (5.171) The particular solution is Zp(k, l) 1 (k² +l+1) E, E2 + 2 1 1 -(k² +l+1) 31+1/3(A1 + A2 + A¡A2) = 1/3[1-1/3(A1 + A2 + A¡A2) +1/o(△? + △2 + Δ우스2 + 2△1△2 + 2A{△2 +2Δ1Δ2)+.](k2 + l+ 1) = 1/27(9k? – 6k + 9l + 5). (5.172) The homogeneous equation (E1, E2 + 2)ž½(k, l) = 0 (5.173) has the solution zh(k, l) = (-2)* f (e – k). (5.174) Therefore, the general solution of equation (5.170) is Zh(k, l) = zh(k, e) + zp(k, l) (5.175) = (-2)* f(l – k) + /27(9k² – 6k + 9l + 5).

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