1.3.4 Example D The function Yk = (c1 + c2k)2*, (1.59) where ci 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 = 4c¡2* + 4c2(k +2)2*. (1.60) %3D (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 = (1.62) 2k+2 (Yk+2 – 2yk+1), 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 = 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).
1.3.4 Example D The function Yk = (c1 + c2k)2*, (1.59) where ci 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 = 4c¡2* + 4c2(k +2)2*. (1.60) %3D (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 = (1.62) 2k+2 (Yk+2 – 2yk+1), 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 = 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).
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 this plz
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 = 4c¡2* + 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
c2k2* = (k/4)(yk+2 – 2yk+1).
(1.63)
10
Difference Equations
Substituting this last result into equation (1.60) gives
c12* = 1/2yk+1 – 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).
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