Proof.If 1 is even and k, o are odd positive integers, then Xn = Xp-1 and xn+1 = Xn–k = Xn-o• that It follows from Eq.(1) bP P=(A+C) Q+ (B+D) P – (30) - (eQ– dP)' and bQ Q= (A+C) P+(B+D) Q – (31) (е Р- d@)* Consequently, we get b P+Q= (32) [d (1– (B+D)) – e (A+C)]' where d (1– (B+ D)) – e (A+C) > 0, е eb (A+C) PQ= (e+d) [K2+(A+ C)] [d K2 – e (A+C)]²" (33)
Proof.If 1 is even and k, o are odd positive integers, then Xn = Xp-1 and xn+1 = Xn–k = Xn-o• that It follows from Eq.(1) bP P=(A+C) Q+ (B+D) P – (30) - (eQ– dP)' and bQ Q= (A+C) P+(B+D) Q – (31) (е Р- d@)* Consequently, we get b P+Q= (32) [d (1– (B+D)) – e (A+C)]' where d (1– (B+ D)) – e (A+C) > 0, е eb (A+C) PQ= (e+d) [K2+(A+ C)] [d K2 – e (A+C)]²" (33)
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
Show me the steps of determine yellow and inf is here
![Theorem 10.If 1 is even and k, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(A+C) (3e– d) < (e+d) (1– (B+ D)),
(29)
is
valid,
provided
(B+D)
1
and
d (1– (B+ D)) – e (A+ C) > 0.
-
Proof.If 1 is even and k, o are odd positive integers, then
Xn = Xp-1 and xn+1
that
Xn-k= Xn-o.
follows from Eq.(1)
bP
P= (A+C) Q+(B+ D) P –
(30
(e Q- dP)'
and
bQ
Q= (A+ C) P+(B+D) Q –
(31)
(е Р- d)*
Consequently, we get
b
P+Q=
(32)
[d (1– (B+ D)) – e (A+C)]'
where d (1– (B+ D)) – e (A+C) > 0,
e b (A+C)
(e+d) [K2+(A+C)][d K2 – e (A+C)]²
е
PQ =
(33)
where K2
Substituting (32) and (33) into (28), we get the condition
(29). Thus, the proof is now completed.O
(1 – (B+ D)), provided (B+ D) < 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F1caac85c-f308-41ae-8ebb-c6e2aeb6fe39%2Ffes43v7_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
(A+C) (3e– d) < (e+d) (1– (B+ D)),
(29)
is
valid,
provided
(B+D)
1
and
d (1– (B+ D)) – e (A+ C) > 0.
-
Proof.If 1 is even and k, o are odd positive integers, then
Xn = Xp-1 and xn+1
that
Xn-k= Xn-o.
follows from Eq.(1)
bP
P= (A+C) Q+(B+ D) P –
(30
(e Q- dP)'
and
bQ
Q= (A+ C) P+(B+D) Q –
(31)
(е Р- d)*
Consequently, we get
b
P+Q=
(32)
[d (1– (B+ D)) – e (A+C)]'
where d (1– (B+ D)) – e (A+C) > 0,
e b (A+C)
(e+d) [K2+(A+C)][d K2 – e (A+C)]²
е
PQ =
(33)
where K2
Substituting (32) and (33) into (28), we get the condition
(29). Thus, the proof is now completed.O
(1 – (B+ D)), provided (B+ D) < 1.
![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%2F1caac85c-f308-41ae-8ebb-c6e2aeb6fe39%2Fwjq9vq_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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