Consider the equation Yk+2 – 6yk+1 + 8yk = 2+ 3k2 – 5 - 3k. (4.146) The characteristic equation is p2 – 6r + 8 = (r – 2)(r – 4) = 0, (4.147) %3D which leads to the following solution of the homogeneous equation: H) = c12* + c24*, (4.148) where c and c2 are arbitrary constants. The families of the terms in R are 2 - (1], k? - [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) %3D 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 – D3* = 2+ 3k2 – 5- 3*. (4.151)
Consider the equation Yk+2 – 6yk+1 + 8yk = 2+ 3k2 – 5 - 3k. (4.146) The characteristic equation is p2 – 6r + 8 = (r – 2)(r – 4) = 0, (4.147) %3D which leads to the following solution of the homogeneous equation: H) = c12* + c24*, (4.148) where c and c2 are arbitrary constants. The families of the terms in R are 2 - (1], k? - [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) %3D 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 – D3* = 2+ 3k2 – 5- 3*. (4.151)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Explain the determine red and all theorem is there
Transcribed Image Text:Case I
None of the roots in the set {ri} occurs in the set {s;}.
For this case, equation (4.135) is the general solution of the tth-order
equation
g(E)yk = 0.
(4.136)
The functions v", i = 1,2, ..., t, are those that appear in Rk plus all those
of the (finite) families of the individual terms that compose R.
134
Difference Equations
Case II
Some of the roots in the set {ri} occur in the set {s;}.
In this situation, the set {ri}+{s;}
ity than the two individual sets of roots. To proceed, determine the general
solution of equation (4.131a), drop all the functions y that appear in y,
and use the remaining functions to find the proper form for the particular
solutions.
now contains roots of higher multiplic-
This procedure for obtaining particular solutions to the inhomogeneous
equation (4.125) can be summarized as follows:
(i) Construct the family of Rk.
(ii) If the family contains no terms of the homogeneous solution, then write
the particular solution y
of that family. Determine the constants of combination such that the
inhomogeneous difference equation is identically satisfied.
(P)
as a linear combination of the members
(iii) If the family contains terms of the homogeneous solution, then multiply
each member of the family by the smallest integral power of k for which
all such terms are removed. The particular solution y can then be
written as a linear combination of the members of this modified family.
Again, determine the constants of combination such that the inhomoge-
neous difference equation is identically satisfied.
The following examples will illustrate the use of the method of undeter-
mined coefficients.
First, assume that g(E) is a nullifying operator of R on the right-hand
side of equation (4.125). Applying g(E) to both sides of equation (4.125) gives
g(E)f(E)yk = 0,
(4.131a)
%3D
wbore
Yk+n
a1Yk+n–1 + + anyk =
Rk,
an #0,
(4.125)
![4.5.2 Example B
Consider the equation
Yk+2 – 6yk+1 + 8yk
2+ 3k2 – 5. 3k.
(4.146)
The characteristic equation is
p2 – 6r + 8 = (r – 2)(r – 4) = 0,
(4.147)
which leads to the following solution of the homogeneous equation:
(H)
Y = c12* + c24*,
(4.148)
where c and c2 are arbitrary constants. The families of the terms in R are
- [1],
k? - [1, k, k?],
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)
y = A+ Bk + Ck² + D3k,
(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 + 3Ck? – D3k = 2 + 3k2 – 5 · 3k.
(4.151)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F061cf21f-b382-44af-92d5-7b23fba381cb%2Fddb2d7d3-186d-4cc6-bfed-215847be9442%2Fhhifeb9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4.5.2 Example B
Consider the equation
Yk+2 – 6yk+1 + 8yk
2+ 3k2 – 5. 3k.
(4.146)
The characteristic equation is
p2 – 6r + 8 = (r – 2)(r – 4) = 0,
(4.147)
which leads to the following solution of the homogeneous equation:
(H)
Y = c12* + c24*,
(4.148)
where c and c2 are arbitrary constants. The families of the terms in R are
- [1],
k? - [1, k, k?],
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)
y = A+ Bk + Ck² + D3k,
(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 + 3Ck? – D3k = 2 + 3k2 – 5 · 3k.
(4.151)
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