LINEAR DIFFERENCE EQUATIONS Example C The second-order, inhomogeneous equation (k +4)yk+2+ Yk+1 – (k + 1)yk = 1 has the following two solutions:
LINEAR DIFFERENCE EQUATIONS Example C The second-order, inhomogeneous equation (k +4)yk+2+ Yk+1 – (k + 1)yk = 1 has the following two solutions:
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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Explain the determaine
![LINEAR DIFFERENCE EQUATIONS
97
Example C
The second-order, inhomogeneous equation
(k + 4)yk+2 + Yk+1 – (k + 1)yk
1
(3.97)
has the following two solutions:
(1)
1
(3.98)
(k + 1)(k+2)’
k+1(2k + 3)
(2)
Yk
(-1)*
4(k + 1)(k + 2)
(3.99)
to its associated homogeneous equation
( (k + 4)yk+2+ Yk+1 –
(k + 1)yk = 0.
(3.100)
These functions have the Casoratian
k+1
L (k +1) :
(3.101)
(k + 2)(k + 3)(k + 4)
The particular solution takes the form
(1)
Yk = c1(k)y + c2(k)y.
(3.102)
= C1
Direct calculation shows that c1(k) and c2 (k) satisfy the equations
Acı (k) = /4(2k +5),
*Ac2(k) = (-1)*+1.
(3.103)
Summing these expressions gives
k
C1 (k) = 14 > (2i + 5) + c1 = 1/4(k +1)² + c1
(3.104)
i=0
and
k
c2(k)
E(-1)* = –1/½[1+ (–1)*]+ c2,
(3.105)
i=0
where ci and c2 are arbitrary constants. Substituting equations (3.98), (3.99),
(3.104), and (3.105) into equation (3.102) and dropping the terms that contain
the arbitrary constants gives
(2k + 3)[1+(-1)*]
8(k + 1)(k + 2)
k +1
(3.106)
4(k + 2)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F81f1f517-d837-45e6-9267-1c02eb87ba4c%2Fb683928b-4eb6-4882-9d47-f47b5646e995%2Fr5d4b2k_processed.jpeg&w=3840&q=75)
Transcribed Image Text:LINEAR DIFFERENCE EQUATIONS
97
Example C
The second-order, inhomogeneous equation
(k + 4)yk+2 + Yk+1 – (k + 1)yk
1
(3.97)
has the following two solutions:
(1)
1
(3.98)
(k + 1)(k+2)’
k+1(2k + 3)
(2)
Yk
(-1)*
4(k + 1)(k + 2)
(3.99)
to its associated homogeneous equation
( (k + 4)yk+2+ Yk+1 –
(k + 1)yk = 0.
(3.100)
These functions have the Casoratian
k+1
L (k +1) :
(3.101)
(k + 2)(k + 3)(k + 4)
The particular solution takes the form
(1)
Yk = c1(k)y + c2(k)y.
(3.102)
= C1
Direct calculation shows that c1(k) and c2 (k) satisfy the equations
Acı (k) = /4(2k +5),
*Ac2(k) = (-1)*+1.
(3.103)
Summing these expressions gives
k
C1 (k) = 14 > (2i + 5) + c1 = 1/4(k +1)² + c1
(3.104)
i=0
and
k
c2(k)
E(-1)* = –1/½[1+ (–1)*]+ c2,
(3.105)
i=0
where ci and c2 are arbitrary constants. Substituting equations (3.98), (3.99),
(3.104), and (3.105) into equation (3.102) and dropping the terms that contain
the arbitrary constants gives
(2k + 3)[1+(-1)*]
8(k + 1)(k + 2)
k +1
(3.106)
4(k + 2)
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