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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 J-Σ(1-i+ 2ι) -Σ (1)-Σί+2 Σ i=1 i=1 i=1 (2.63 = (k – 1) – k(k – 1) , (k – 1)²k² The general solution is Yk = 1/½k4 – k³ + 3/½k + A, (2.64 where A is an arbitrary constant. In terms of Bernoulli polynomials, this las expression reads Yk B1(k) – 1/½B2(k) + /½B4(k) + A1, (2.65 IWI

2.3.1 Example The equation Yk+1 – Yk = 1 – k+ 2k3 (2.62 has the particular solution k-1 k-1 k-1 J-Σ(1-i+ 2ι) -Σ (1)-Σί+2 Σ i=1 i=1 i=1 (2.63 = (k – 1) – k(k – 1) , (k – 1)²k² The general solution is Yk = 1/½k4 – k³ + 3/½k + A, (2.64 where A is an arbitrary constant. In terms of Bernoulli polynomials, this las expression reads Yk B1(k) – 1/½B2(k) + /½B4(k) + A1, (2.65 IWI

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 determaine tis step

FIRST-ORDER DIFFERENCE EQUATIONS 53 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 υ-Σ1-ί+ 28) Σ (1) - Σί+2Σ i=1 (2.63) i=1 i=1 i=1 = (k – 1) k(k – 1) + (k – 1)²k² 2 The general solution is Yk = 1/½k4 – k³ + 3/2k + A, (2.64) where A is an arbitrary constant. In terms of Bernoulli polynomials, this last expression reads Yk B1(k) – 1/½B2(k) + /½B4(k) + A1, (2.65) where A1 is an arbitrary function. The result of equation (2.65) could have been immediately written down by first noting that equation (2.60) is a linear equation and thus its particular solution is a sum of particular solutions of equations having the form Yk+1 – Yk = amk", 0<m<n. (2.66) Under the transformation Am Вт+1(k), т +1 (2.67) Yk equation (2.66) becomes Bm+1(k+1) – Bm+1(k) = (m + 1)k", (2.68) which is the defining difference equation for the Bernoulli polynomials. There- fore, from the known coefficients (am), the particular solution can be imme- diately obtained from equation (2.67).
Transcribed Image Text:FIRST-ORDER DIFFERENCE EQUATIONS 53 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 υ-Σ1-ί+ 28) Σ (1) - Σί+2Σ i=1 (2.63) i=1 i=1 i=1 = (k – 1) k(k – 1) + (k – 1)²k² 2 The general solution is Yk = 1/½k4 – k³ + 3/2k + A, (2.64) where A is an arbitrary constant. In terms of Bernoulli polynomials, this last expression reads Yk B1(k) – 1/½B2(k) + /½B4(k) + A1, (2.65) where A1 is an arbitrary function. The result of equation (2.65) could have been immediately written down by first noting that equation (2.60) is a linear equation and thus its particular solution is a sum of particular solutions of equations having the form Yk+1 – Yk = amk", 0<m<n. (2.66) Under the transformation Am Вт+1(k), т +1 (2.67) Yk equation (2.66) becomes Bm+1(k+1) – Bm+1(k) = (m + 1)k", (2.68) which is the defining difference equation for the Bernoulli polynomials. There- fore, from the known coefficients (am), the particular solution can be imme- diately obtained from equation (2.67).
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