2.10. Solve the following equations: a) Yk+1 = (k² +1)yk,

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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2.3.1
Example
The equation
Yk+1 – Yk =1 – k+ 2k³
(2.62)
has the particular solution
k-1
k-1
k-1
k-1
E(1 - i+ 2i³)
-Σ1)-Σi+2Σ
Yk
i=1
i=1
i=1
i=1
(2.63)
k(k – 1)
(k – 1)²k²
= (k – 1) .
2
The general solution is
Yk = /½k4 – k° + 3½k + A,
(2.64)
where A is an arbitrary constant. In terms of Bernoulli polynomials, this last
expression reads
Yk = B1(k) – 1/2B2(k) + /½B4(k)+ A1,
(2.65)
Transcribed Image Text:2.3.1 Example The equation Yk+1 – Yk =1 – k+ 2k³ (2.62) has the particular solution k-1 k-1 k-1 k-1 E(1 - i+ 2i³) -Σ1)-Σi+2Σ Yk i=1 i=1 i=1 i=1 (2.63) k(k – 1) (k – 1)²k² = (k – 1) . 2 The general solution is Yk = /½k4 – k° + 3½k + A, (2.64) where A is an arbitrary constant. In terms of Bernoulli polynomials, this last expression reads Yk = B1(k) – 1/2B2(k) + /½B4(k)+ A1, (2.65)
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
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