Example B Consider the two linearly independent functions (1) (2) Y = 1, y = 3k. A second-order equation having these solutions is (4.96) 3k 1 3k+1 1 3k+2 Ук 1 1 1 3k Yk+1 Yk 1 Yk+1 3 0, (4.97) Yk+2 Yk+2 1 9. or Yk+2 – 4yk+1 +3yk 0. (4.98)

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