(PQ) [(A+D+1) − (B+C)] = 0, -
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 green and the inf is here
![Theorem 8. If o is even and k, 1 are odd positive integers
and (A+D+ 1) + (B+ C), then Eq.(1) has no prime
period two solution.
Proof.Following the proof of Theorem 5, we deduce that if
o is even and k, 1 are odd positive integers, then x, = Xn-o
and Xn+1 = xn-k= Xn=1. It follows from Eq.(1) that
b
P=(A+D) Q+(B+C)P –
(18)
(e – d)
and
b
Q = (A+D) P+ (B+C) Q –
(19)
(e – d)
By subtracting (19) from (18), we get
(P- Q) [(A+D+ 1) – (B+C)] = 0,
Since (A+D+1) – (B+C) + 0, then P= Q. This is a
contradiction. Thus, the proof is now completed.O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb19240c-b74e-434f-b44c-0f0733785fa6%2F163a902e-4438-4052-aa65-ac6255a25d36%2Fkfdkzd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 8. If o is even and k, 1 are odd positive integers
and (A+D+ 1) + (B+ C), then Eq.(1) has no prime
period two solution.
Proof.Following the proof of Theorem 5, we deduce that if
o is even and k, 1 are odd positive integers, then x, = Xn-o
and Xn+1 = xn-k= Xn=1. It follows from Eq.(1) that
b
P=(A+D) Q+(B+C)P –
(18)
(e – d)
and
b
Q = (A+D) P+ (B+C) Q –
(19)
(e – d)
By subtracting (19) from (18), we get
(P- Q) [(A+D+ 1) – (B+C)] = 0,
Since (A+D+1) – (B+C) + 0, then P= Q. This is a
contradiction. Thus, the proof is now completed.O
![Theorem 5.If k,1 and o are both even positive integers,
then Eq. (1) has no prime period two solution.
Proof.Assume that there exists distinct positive solutions
P,Q, P, Q,..
of prime period two of Eq.(1). If k,1 and o are both even
positive integers, then x, = Xp-k= Xn-1= Xn-o.
from Eq. (1) that
It follows
P=(A+B+C+D) Q –
(e
d)
(12)
|
and
b
Q= (A+B+C+D)P –
(13)
(e – d)'
|
By subtracting (13) from (12), we get
(P- Q) [A+B+C+D+1]=0.
||
Since A+ B+C +D+1 0, then P:
contradiction. Thus, the proof is now completed.O
Q. This is a
6.
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxr- k
[dxn-k- exn-1]
(1)
Xn+1
Axn+ Bxn–k+ Cxn–1+Dxn-o +
n = 0,1, 2, ....
where the coefficients A, B, C, D, b, d, e € (0,0), while
k,1 and o are positive integers. The initial conditions
X-,..., X_1,..., X_k ….., X_1, Xo are arbitrary positive real
numbers such that k <1< 0. Note that the special cases
of Eq. (1) have been studied in [1] when B=C= D=0,
and k= 0,1= 1, b is replaced by – b and in [27] when
B=C= D=0, and k= 0, b is replaced by – b and in
[33] when B = C = D = 0, 1= 0 and in [32] when
A=C= D=0, 1=0, b is replaced by – b.
6.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb19240c-b74e-434f-b44c-0f0733785fa6%2F163a902e-4438-4052-aa65-ac6255a25d36%2F4yr31hr_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 5.If k,1 and o are both even positive integers,
then Eq. (1) has no prime period two solution.
Proof.Assume that there exists distinct positive solutions
P,Q, P, Q,..
of prime period two of Eq.(1). If k,1 and o are both even
positive integers, then x, = Xp-k= Xn-1= Xn-o.
from Eq. (1) that
It follows
P=(A+B+C+D) Q –
(e
d)
(12)
|
and
b
Q= (A+B+C+D)P –
(13)
(e – d)'
|
By subtracting (13) from (12), we get
(P- Q) [A+B+C+D+1]=0.
||
Since A+ B+C +D+1 0, then P:
contradiction. Thus, the proof is now completed.O
Q. This is a
6.
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxr- k
[dxn-k- exn-1]
(1)
Xn+1
Axn+ Bxn–k+ Cxn–1+Dxn-o +
n = 0,1, 2, ....
where the coefficients A, B, C, D, b, d, e € (0,0), while
k,1 and o are positive integers. The initial conditions
X-,..., X_1,..., X_k ….., X_1, Xo are arbitrary positive real
numbers such that k <1< 0. Note that the special cases
of Eq. (1) have been studied in [1] when B=C= D=0,
and k= 0,1= 1, b is replaced by – b and in [27] when
B=C= D=0, and k= 0, b is replaced by – b and in
[33] when B = C = D = 0, 1= 0 and in [32] when
A=C= D=0, 1=0, b is replaced by – b.
6.
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