2.10. Solve the following equations: (a) Yk+1 = (k² + 1)yk, (b) Yk+1 = (k – k² – k³)yk, (c) = . - - Yk+1 k+1 Yk k

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.3: Quadratic Equations
Problem 51E
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2.10. Solve the following equations:
(k² + 1)yk,
= (k – k? – k*)yk,
(a) Yk+1
(b) Yk+1
-
(c) = .
Yk+1
k+1
Yk
k
Transcribed Image Text:2.10. Solve the following equations: (k² + 1)yk, = (k – k? – k*)yk, (a) Yk+1 (b) Yk+1 - (c) = . Yk+1 k+1 Yk k
2.4.1
Example A
The equation
Yk+1 = (k – k²)Yk
(2.73)
can be written as
Yk+1 = (-)k(k – 1)yk-
(2.74)
Its solution is
= A(-)*T(k)F(k – 1) = A(-)*(k – 1)r°(k – 1),
(2.75)
Yk =
where A is an arbitrary constant.
Transcribed Image Text:2.4.1 Example A The equation Yk+1 = (k – k²)Yk (2.73) can be written as Yk+1 = (-)k(k – 1)yk- (2.74) Its solution is = A(-)*T(k)F(k – 1) = A(-)*(k – 1)r°(k – 1), (2.75) Yk = where A is an arbitrary constant.
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