The combined family is [1, k, k², 3k] and contains no members that occur in the homogeneous solution. Therefore, the particular solution takes the form (P) Yk = A+ Bk + Ck2 + D3*, (4.150) where A, B, C, and D are constants to be determined. Substitution of equation (4.150) into (4.146) and simplifying the resulting expression gives (3A – 4B – 2C) + (3B – 8C')k + 3Ck² – D3k = 2+ 3k? – 5- 3*. (4.151)

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

Example B
Consider the equation
Yk+2
6yk+1 + 8yk = 2+ 3k2 – 5 · 3k.
(4.146)
The characteristic equation is
p? – 6r + 8 = (r – 2)(r – 4) = 0,
(4.147)
which leads to the following solution of the homogeneous equation:
(H)
c12k + c24*,
(4.148)
where c, and c2 are arbitrary constants. The families of the terms in Rp are
[1],
k² → [1, k, k²],
3k → [3*).
(4.149)
The combined family is [1, k, k² , 3k] and contains no members that occur in
the homogeneous solution. Therefore, the particular solution takes the form
(P)
A + Bk + Ck2 + D3*,
(4.150)
where A, B, C, and D are constants to be determined. Substitution of equation
(4.150) into (4.146) and simplifying the resulting expression gives
(ЗА
4B - 2С) + (3В — 8C)k + ЗСk? — D3*
= 2+ 3k2 – 5 . 3k. (4.151)
136
Difference Equations
Equating the coefficients of the linearly independent terms on both sides to
zero gives
ЗА - 4В — 2С — 2,
ЗВ - 8C %3 0, ЗС %3D 3, D3 5,
(4.152)
%3|
which has the solution
A = 44/9,
B = 8/3, C=1,
D = 5.
(4.153)
Consequently, the particular solution is
= 44/9 + 8/3k + k² + 5 • 3*,
(4.154)
and the general solution to equation (4.146) is
Yk = C12k + c24% + 44/9 + 8/3k + k² + 5 · 3*.
(4.155)
Transcribed Image Text:Example B Consider the equation Yk+2 6yk+1 + 8yk = 2+ 3k2 – 5 · 3k. (4.146) The characteristic equation is p? – 6r + 8 = (r – 2)(r – 4) = 0, (4.147) which leads to the following solution of the homogeneous equation: (H) c12k + c24*, (4.148) where c, and c2 are arbitrary constants. The families of the terms in Rp are [1], k² → [1, k, k²], 3k → [3*). (4.149) The combined family is [1, k, k² , 3k] and contains no members that occur in the homogeneous solution. Therefore, the particular solution takes the form (P) A + Bk + Ck2 + D3*, (4.150) where A, B, C, and D are constants to be determined. Substitution of equation (4.150) into (4.146) and simplifying the resulting expression gives (ЗА 4B - 2С) + (3В — 8C)k + ЗСk? — D3* = 2+ 3k2 – 5 . 3k. (4.151) 136 Difference Equations Equating the coefficients of the linearly independent terms on both sides to zero gives ЗА - 4В — 2С — 2, ЗВ - 8C %3 0, ЗС %3D 3, D3 5, (4.152) %3| which has the solution A = 44/9, B = 8/3, C=1, D = 5. (4.153) Consequently, the particular solution is = 44/9 + 8/3k + k² + 5 • 3*, (4.154) and the general solution to equation (4.146) is Yk = C12k + c24% + 44/9 + 8/3k + k² + 5 · 3*. (4.155)
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