Note that equation (2.29) corresponds to having pk = k/(k+1). Therefore, %3D k-1 k-1 i (k – 1)! k! 1 II P: = II (2.34) i +1 i=1 k i=1

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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FIRST-ORDER DIFFERENCE EQUATIONS
49
2.2.5 Example E
The equation
(k + 1)yk+1 – kyk = 0
(2.29)
can be put in simpler form by the transformation
Zk = kyk,
(2.30)
which gives
Zk+1
- Zk = 0.
(2.31)
The solution to this latter equation is
Zk = A = arbitrary constant.
(2.32)
Therefore,
Yk =
= A/k.
(2.33)
Note that equation (2.29) corresponds to having pk = k/(k+1). Therefore,
k-1
k-
(k – 1)!
1
Il Pi
(2.34)
+1
k!
k
i=1
i=1
Putting this result into equation (2.4) again gives the solution of equation
(2.33).
Transcribed Image Text:FIRST-ORDER DIFFERENCE EQUATIONS 49 2.2.5 Example E The equation (k + 1)yk+1 – kyk = 0 (2.29) can be put in simpler form by the transformation Zk = kyk, (2.30) which gives Zk+1 - Zk = 0. (2.31) The solution to this latter equation is Zk = A = arbitrary constant. (2.32) Therefore, Yk = = A/k. (2.33) Note that equation (2.29) corresponds to having pk = k/(k+1). Therefore, k-1 k- (k – 1)! 1 Il Pi (2.34) +1 k! k i=1 i=1 Putting this result into equation (2.4) again gives the solution of equation (2.33).
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