Explain the determaine 1.3.4 Example DThe function= (c1 +c2k)2*,33(1.59,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 = 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+2orczk2* = (k/4)(yk+2 – 2yk+1).(1.63)10Difference EquationsSubstituting this last result into equation (1.60) givesc12* = /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 isobtained:Yk+2 – 4Yk+1+ 4yk = 0.(1.65)This is the difference equation whose solution is equation (1.59).

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Description: Explain the determaine 1.3.4 Example DThe function= (c1 +c2k)2*,33(1.59,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 +

Given function is yk=(c1+c2k)2k (1) Now applying the difference operator ∆ in (1) We get…

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1.3.4 Example D The function Yk = (c1 + c2k)2*, (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*, Yk+2 = 4c12* + 4c2(k+2)2*. (1.60) (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 C2 = (Yk+2 – 2yk+1), (1.62) 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 – 4(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).

1.3.4 Example D The function Yk = (c1 + c2k)2, (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, Yk+2 = 4c12 + 4c2(k+2)2. (1.60) (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 C2 = (Yk+2 – 2yk+1), (1.62) 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 – 4(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).

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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Explain the determaine

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).

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