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).
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 5T
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