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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 Σ(1-i+23) Σ(1) -Σ i+2i %3| i=1 i=1 i=1 (2.63) i=1 k(k – 1) (k – 1)²k² = (k – 1) - 2 The general solution is Yk = 1/2k* – k° + 3/2k + A, (2.64) where A is an arbitrary constant. In terms of Bernoulli polynomials, this last expression reads

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 Σ(1-i+23) Σ(1) -Σ i+2i %3| i=1 i=1 i=1 (2.63) i=1 k(k – 1) (k – 1)²k² = (k – 1) - 2 The general solution is Yk = 1/2k* – k° + 3/2k + A, (2.64) where A is an arbitrary constant. In terms of Bernoulli polynomials, this last expression reads

Advanced Engineering Mathematics
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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Explain the determaine step by step how he was solved the general solution

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 Yk =E(1- i+ 2i*) = (1) –i+2 i=1 i=: (2.63) i=1 i=1 k(k – 1) (k – 1)²k? = (k – 1) 2 The general solution is Yk = /2k* – k° + 3/½k + A, (2.64) where A is an arbitrary constant. In terms of Bernoulli polynomials, this last expression reads Yk = B1(k) – 1/2B2(k)+1/½B4(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), m +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 Yk =E(1- i+ 2i*) = (1) –i+2 i=1 i=: (2.63) i=1 i=1 k(k – 1) (k – 1)²k? = (k – 1) 2 The general solution is Yk = /2k* – k° + 3/½k + A, (2.64) where A is an arbitrary constant. In terms of Bernoulli polynomials, this last expression reads Yk = B1(k) – 1/2B2(k)+1/½B4(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), m +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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