6.5.3 Example C If in the equation (Yk+2)² – 4(yk+1)² +3(yk)² = k (6.83) the substitution xk yi is made, then the following result is obtained: Xk+2 – 4xk+1 +3xk = k. (6.84) The solution to the latter equation is Xk = C1 + c23k – 1/¼k2 (6.85a) and Yk = c1 + c23k – 1/¼k² . (6.85b) Consequently, equation (6.83) has the two solutions (: Yk = +(c1 + c23* – 1/¼k²)!/2 (6.86a) or Yk = -(c1 + c23* – 1/¼k²)!/2 (6.86b)
6.5.3 Example C If in the equation (Yk+2)² – 4(yk+1)² +3(yk)² = k (6.83) the substitution xk yi is made, then the following result is obtained: Xk+2 – 4xk+1 +3xk = k. (6.84) The solution to the latter equation is Xk = C1 + c23k – 1/¼k2 (6.85a) and Yk = c1 + c23k – 1/¼k² . (6.85b) Consequently, equation (6.83) has the two solutions (: Yk = +(c1 + c23* – 1/¼k²)!/2 (6.86a) or Yk = -(c1 + c23* – 1/¼k²)!/2 (6.86b)
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 determine blue
Transcribed Image Text:6.5.3 Example C
If in the equation
(Yk+2)² – 4(yk+1)² + 3(yr)² = k
(6.83)
the substitution xk =
yi is made, then the following result is obtained:
Xk+2 – 4xk+1 + 3xk
k.
(6.84)
The solution to the latter equation is
Xk = c1 + c23k – 1/¼k2
(6.85a)
and
Yk = c1 + c23k – 1¼k2.
(6.85b)
Consequently, equation (6.83) has the two solutions
+(c1 + c23* – 1/4k²)1/2
(6.86a)
Yk =
or
-(cı + c23* – 1/¼k?)!/2,
(6.86b)
Yk
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