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The particular solution is (P) Yk - (Е — 3)-1(Е — 2)-13*. | Note that f(3) = 0; applying Theorem 4.9', we obtain k3k-1 (P) Yk = k3k-1 (3 – 2)(1!) | Therefore, the general solution of equation (4.227) is

The particular solution is (P) Yk - (Е — 3)-1(Е — 2)-13*. | Note that f(3) = 0; applying Theorem 4.9', we obtain k3k-1 (P) Yk = k3k-1 (3 – 2)(1!) | Therefore, the general solution of equation (4.227) is

Advanced Engineering Mathematics
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 red and Theorem is there

Theorem 4.9'. Let f(a) = 0, where f(E) = (E – a)"g(E) and g(a) + 0. Then -m km ak- f(E)-'ak (4.206) g(a)m! Proof. We assume that f(r) has a zero atr=a of multiplicity m. Therefore, ak f(E)-'a* = (E – a)-"g(E)¯'a* = (E – a)* g(a) -1k -m ak - (aЕ — а)- ak-m 1 g(a) ak-m k(m) -(E – 1) ™. 1 (4.207) -m g(a) k-m -m 1 = g(a) g(a) m!
Transcribed Image Text:Theorem 4.9'. Let f(a) = 0, where f(E) = (E – a)"g(E) and g(a) + 0. Then -m km ak- f(E)-'ak (4.206) g(a)m! Proof. We assume that f(r) has a zero atr=a of multiplicity m. Therefore, ak f(E)-'a* = (E – a)-"g(E)¯'a* = (E – a)* g(a) -1k -m ak - (aЕ — а)- ak-m 1 g(a) ak-m k(m) -(E – 1) ™. 1 (4.207) -m g(a) k-m -m 1 = g(a) g(a) m!
Example C The equation Yk+2 – 5yk+1 + 6yk = 3k (4.227) can be expressed as f(E)yk = (E – 2)(E – 3)yk = 3*. (4.228) 144 Difference Equations The particular solution is P) = (E – 3)-'(E – 2)-13k. (Р) (4.229) Note that f(3) = 0; applying Theorem 4.9', we obtain (Р) k3k-1 = k3k-1. (4.230) (3 – 2)(1!) Therefore, the general solution of equation (4.227) is c12* + c23k + k3k-1. (4.231) Yk =
Transcribed Image Text:Example C The equation Yk+2 – 5yk+1 + 6yk = 3k (4.227) can be expressed as f(E)yk = (E – 2)(E – 3)yk = 3*. (4.228) 144 Difference Equations The particular solution is P) = (E – 3)-'(E – 2)-13k. (Р) (4.229) Note that f(3) = 0; applying Theorem 4.9', we obtain (Р) k3k-1 = k3k-1. (4.230) (3 – 2)(1!) Therefore, the general solution of equation (4.227) is c12* + c23k + k3k-1. (4.231) Yk =
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