CP where d (1- (B+D)) -e (A+C) > 0, e b² (A+C) (e+d) [K₂ + (A+C)] [d K₂ −e (A+C)]² ³ PQ= P+Q= b [d (1 − (B+ D)) − e (A+C)]' (32) (33)
CP where d (1- (B+D)) -e (A+C) > 0, e b² (A+C) (e+d) [K₂ + (A+C)] [d K₂ −e (A+C)]² ³ PQ= P+Q= b [d (1 − (B+ D)) − e (A+C)]' (32) (33)
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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Question
Explain the determine green and the information ti here
![The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn– k
X+1 = Axn+ Bxp–k+CXp-1+Dxp-o+
[dxn-k– ex-1
(1)
n= 0, 1,2, .....
where the coefficients A, B, C, D, b, d, e e (0,00), while
k, 1 and o are positive integers. The initial conditions
X-g,..., 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=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.
••..
- b and in [27] when
6.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd6e1725f-ce99-4ac2-8766-635487711a2a%2F109ead8e-e4d7-42d9-8131-f92ebf0a42cc%2Fessyiq9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn– k
X+1 = Axn+ Bxp–k+CXp-1+Dxp-o+
[dxn-k– ex-1
(1)
n= 0, 1,2, .....
where the coefficients A, B, C, D, b, d, e e (0,00), while
k, 1 and o are positive integers. The initial conditions
X-g,..., 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=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.
••..
- b and in [27] when
6.
![Theorem 10. If 1 is even and k, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(А+ C) (Зе— d) < (е+ d) (1-(В+ D),
(29)
is
valid,
(B+ D)
and
provided
d (1 — (В+ D) — е (А+С)>0.
< 1
Proof.If 1 is even and k,
are odd positive integers, then
Xn = Xp–1 and xn+1 = Xn-k = Xn-o. It follows from Eq.(1)
that
bP
Р- (А+C)Q+ (B+D) P —
(30)
(e Q– dP)'
and
bQ
Q= (A+C) P+(B+D) Q –
(31)
(e P- dQ)'
Consequently, we get
b
P+Q=
(32)
[d (1 — (В+ D)) — е (А+С)]"
where d (1– (B+D)) – e (A+C) > 0,
e b² (A+C)
PQ=
(e+d) [K2+(A+ C)][d K2 – e (A+C)]²"
(33)
-
where K2
(1– (B+D)), provided (B+ D) < 1.
Substituting (32) and (33) into (28), we get the condition
(29). Thus, the proof is now completed.O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd6e1725f-ce99-4ac2-8766-635487711a2a%2F109ead8e-e4d7-42d9-8131-f92ebf0a42cc%2F2vqfbap_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 10. If 1 is even and k, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(А+ C) (Зе— d) < (е+ d) (1-(В+ D),
(29)
is
valid,
(B+ D)
and
provided
d (1 — (В+ D) — е (А+С)>0.
< 1
Proof.If 1 is even and k,
are odd positive integers, then
Xn = Xp–1 and xn+1 = Xn-k = Xn-o. It follows from Eq.(1)
that
bP
Р- (А+C)Q+ (B+D) P —
(30)
(e Q– dP)'
and
bQ
Q= (A+C) P+(B+D) Q –
(31)
(e P- dQ)'
Consequently, we get
b
P+Q=
(32)
[d (1 — (В+ D)) — е (А+С)]"
where d (1– (B+D)) – e (A+C) > 0,
e b² (A+C)
PQ=
(e+d) [K2+(A+ C)][d K2 – e (A+C)]²"
(33)
-
where K2
(1– (B+D)), provided (B+ D) < 1.
Substituting (32) and (33) into (28), we get the condition
(29). Thus, the proof is now completed.O
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