1.3.4 Example D The function = (c1 +c2k)2*, 33 (1.59, where c1 and c2 are arbitrary constants, is the solution to a second-order difference equation. To determine this equation, we first calculate...

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Extracted text: 1.3.4 Example D The function = (c1 +c2k)2*, 33 (1.59, where c1 and c2 are arbitrary constants, is the solution to a second-order difference equation. To determine this equation, we first calculate yk+1 and Yk+2: Yk+1 = 2c12* + 2c2(k + 1)2*, (1.60) Yk+2 = 4c12* + 4c2(k + 2)2*. (1.61) Multiplying equation (1.60) by 2 and subtracting this from equation (1.61) gives an expression that can be solved for c2: 1 (Yk+2 – 2yk+1), (1.62) C2 = 2k+2 or czk2* = (k/4)(yk+2 – 2yk+1). (1.63) 10 Difference Equations Substituting this last result into equation (1.60) gives c12* = /2yk+1 – ¼(k + 1)(Yk+2 – 2yk+1). (1.64) If equations (1.63) and (1.64) are used in the right-hand side of equation (1.59) and if the resulting expression is simplified, then the following result is obtained: Yk+2 – 4Yk+1+ 4yk = 0. (1.65) This is the difference equation whose solution is equation (1.59).