6.5.2 Example B The equation (k +1)y+1 – ky = 0 (6.79) can be transformed to the linear form Xk+1 – k = 0 (6.80) by means of the substitution xk = ky?. Since xk = c is the solution of equation (6.80), we obtain (kyk = c (6.81) and Yk = +Vc/k_or Yk = – Vc/k, (6.82) as two solutions of equation (6.79).
6.5.2 Example B The equation (k +1)y+1 – ky = 0 (6.79) can be transformed to the linear form Xk+1 – k = 0 (6.80) by means of the substitution xk = ky?. Since xk = c is the solution of equation (6.80), we obtain (kyk = c (6.81) and Yk = +Vc/k_or Yk = – Vc/k, (6.82) as two solutions of equation (6.79).
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:6.5.2 Example B
The equation
(k + 1)y+1 – kyj = 0
(6.79)
can be transformed to the linear form
Xk+1 – Xk = 0
(6.80)
by means of the substitution x = ky?. Since x = c is the solution of equation
(6.80), we obtain
kyr
(6.81)
= C
and
Yk = +Vc/k_or
Yk = - Vc/k,
(6.82)
as two solutions of equation (6.79).
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