Explain the determaine step by step how he was solved the general solution FIRST-ORDER DIFFERENCE EQUATIONS532.3.1ExampleThe equationYk+1 – Yk = 1 – k + 2k3(2.62)has the particular solutionk-1k-1k-1k-1Yk =E(1- i+ 2i*) = (1) –i+2i=1i=:(2.63)i=1i=1k(k – 1)(k – 1)²k?= (k – 1)2The general solution isYk = /2k* – k° + 3/½k + A,(2.64)where A is an arbitrary constant. In terms of Bernoulli polynomials, this lastexpression readsYk = 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 downby first noting that equation (2.60) is a linear equation and thus its particularsolution is a sum of particular solutions of equations having the formYk+1 – Yk =amk", 0< m< n.(2.66)Under the transformationAmВт+1(k),m +1(2.67)Ykequation (2.66) becomesBm+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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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

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter5: Inverse, Exponential, And Logarithmic Functions

Section5.6: Exponential And Logarithmic Equations

Problem 3E

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