Theorem 11.If 1,0 are even and k is odd positive integers, then Eq.(1) has prime period two solution if the condition (A+C+D)(3e- d) < (e+d)(1 – B), (34) | is valid, provided B< 1 and d (1 – B) – e (A+C+D) > 0. | Proof.If 1,0 are even and k is odd positive integers, then Xn = Xn-1= Xn-o and xn+1 = Xn=k• It follows from Eq.(1) that bP P= (A+C+D) Q+BP (35) %3| (e Q- dP)' and bQ Q= (A+C+D) P+BQ (36) (еР — do) Consequently, we get P+Q= (37) [d (1– B) – e (A+C+D)]' where d (1- B) - e (A+C+D) > 0, e (A+C+D) (e+d) [(1– B) + K3] [d (1 – B) – e K]² PQ= (38) (A+C+ D), provided B< 1. Substituting where K3 = (37) and (38) into (28), we get the condition (34). Thus, the proof is now completed.O
Theorem 11.If 1,0 are even and k is odd positive integers, then Eq.(1) has prime period two solution if the condition (A+C+D)(3e- d) < (e+d)(1 – B), (34) | is valid, provided B< 1 and d (1 – B) – e (A+C+D) > 0. | Proof.If 1,0 are even and k is odd positive integers, then Xn = Xn-1= Xn-o and xn+1 = Xn=k• It follows from Eq.(1) that bP P= (A+C+D) Q+BP (35) %3| (e Q- dP)' and bQ Q= (A+C+D) P+BQ (36) (еР — do) Consequently, we get P+Q= (37) [d (1– B) – e (A+C+D)]' where d (1- B) - e (A+C+D) > 0, e (A+C+D) (e+d) [(1– B) + K3] [d (1 – B) – e K]² PQ= (38) (A+C+ D), provided B< 1. Substituting where K3 = (37) and (38) into (28), we get the condition (34). Thus, the proof is now completed.O
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
How to deduce this equation from Equation 1 Explain to me the method. Show me the steps of determine blue and equation 1 is second pic
![Theorem 11.If 1,0
are even and k is odd positive
integers, then Eq.(1) has prime period two solution if the
condition
(А+С+D) (Зе— d) < (е+ d) (1—В),
(34)
is valid, provided B< 1 and d (1 – B) – e (A+C+D) >
0.
Proof.If 1,0 are even and k is odd positive integers, then
Xn = Xn-1= Xn-o and xn+1 = Xp-k. It follows from Eq.(1)
that
bP
P=(A+C+D)Q+BP
(35)
(e Q– dP)’
and
Q= (A+C+ D) P+BQ
bQ
(еР — d0)"
(36)
Consequently, we get
b
P+Q=
(37)
[d (1– B) – e (A+C+D)]'
where d (1- B) - e (A+C+D) > 0,
eb (A+C+D)
(e+d) [(1– B) + K3] [d (1– B) – e K3]?
PQ=
(38)
where K3
(37) and (38) into (28), we get the condition (34). Thus,
the proof is now completed.O
(A+C+D), provided B< 1. Substituting](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F81e48d53-799e-4b93-8de4-a9263a6f3712%2F1d7c5bz_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 11.If 1,0
are even and k is odd positive
integers, then Eq.(1) has prime period two solution if the
condition
(А+С+D) (Зе— d) < (е+ d) (1—В),
(34)
is valid, provided B< 1 and d (1 – B) – e (A+C+D) >
0.
Proof.If 1,0 are even and k is odd positive integers, then
Xn = Xn-1= Xn-o and xn+1 = Xp-k. It follows from Eq.(1)
that
bP
P=(A+C+D)Q+BP
(35)
(e Q– dP)’
and
Q= (A+C+ D) P+BQ
bQ
(еР — d0)"
(36)
Consequently, we get
b
P+Q=
(37)
[d (1– B) – e (A+C+D)]'
where d (1- B) - e (A+C+D) > 0,
eb (A+C+D)
(e+d) [(1– B) + K3] [d (1– B) – e K3]?
PQ=
(38)
where K3
(37) and (38) into (28), we get the condition (34). Thus,
the proof is now completed.O
(A+C+D), provided B< 1. Substituting
![Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxn-k - exp-1)
Xn+1 =
Axn+ Bxn-k+Cxn–1+Dxn-o +
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-6…, X_1,..., X_k, ..., X_1, Xo are arbitrary positive real
numbers such that k <1< o. 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
[33] when B = C = D = 0, 1 = 0 and in [32] when
A = C= D=0, 1= 0, b is replaced by – b.
(1)
b and in
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F81e48d53-799e-4b93-8de4-a9263a6f3712%2F89m59au_processed.png&w=3840&q=75)
Transcribed Image Text:Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
[dxn-k - exp-1)
Xn+1 =
Axn+ Bxn-k+Cxn–1+Dxn-o +
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-6…, X_1,..., X_k, ..., X_1, Xo are arbitrary positive real
numbers such that k <1< o. 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
[33] when B = C = D = 0, 1 = 0 and in [32] when
A = C= D=0, 1= 0, b is replaced by – b.
(1)
b and in
%3D
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