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gives an expression that can be solved for c2: 1 C2 = (Yk+2 -2yk+1), (1.62) 2k+2 or czk2* = (k/4)(yk+2 – 2yk+1). (1.63) %3D 10 Difference Equations Substituting this last result into equation (1.60) gives c12* = 1/2yk+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

gives an expression that can be solved for c2: 1 C2 = (Yk+2 -2yk+1), (1.62) 2k+2 or czk2* = (k/4)(yk+2 – 2yk+1). (1.63) %3D 10 Difference Equations Substituting this last result into equation (1.60) gives c12* = 1/2yk+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

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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1.3.4 Example D The function (c1 + c2k)2*, (1.59) Yk = 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 = 4c12k + 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) - 2k+2 or c2k2k = (k/4)(yk+2 – 2yk+1). (1.63) 10 Difference Equations Substituting this last result into equation (1.60) gives c12* = 1/2Yk+1 – 1¼(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).
Transcribed Image Text:1.3.4 Example D The function (c1 + c2k)2*, (1.59) Yk = 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 = 4c12k + 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) - 2k+2 or c2k2k = (k/4)(yk+2 – 2yk+1). (1.63) 10 Difference Equations Substituting this last result into equation (1.60) gives c12* = 1/2Yk+1 – 1¼(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).
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