Explain this Q plz! 1.3.4Example DThe function(c1 + c2k)2*,(1.59)Yk =where c1 and c2 are arbitrary constants, is the solution to a second-orderdifference equation. To determine this equation, we first calculate yk+1 andYk+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+2orc2k2k = (k/4)(yk+2 – 2yk+1).(1.63)10Difference EquationsSubstituting this last result into equation (1.60) givesc12* = 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 isobtained:Yk+2 – 4yk+1 + 4yk= 0.(1.65)|This is the difference equation whose solution is equation (1.59).

Answered: Image /qna-images/answer/c0b0bec0-deef-4422-aef2-a8bb879faf76.jpg

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

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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