Theorem 12.If k, o are even and 1 is odd positive integers, then Eq. (1) has prime period two solution if the condition (3e-d) (1-C) < (e+d) (A+B+D), (39) is valid, provided C < 1 and e (1− C) −d (A+B+D) > 0. Proof. If k,o are even and I is odd positive integers, then Xn = Xn-k = Xn-o and Xn+1 and Xn+1 = Xn-1. It follows from Eq. (1) that and P= (A+B+ D) Q+ CP Q= (A+B+D) P+CQ Consequently, we get - b CP+Q=(1-0)- #1 bQ (e P- dQ)' bP (e Q- dp)* [e (1 − C) −d (A+B+D)]' (40) (41) (42)
Theorem 12.If k, o are even and 1 is odd positive integers, then Eq. (1) has prime period two solution if the condition (3e-d) (1-C) < (e+d) (A+B+D), (39) is valid, provided C < 1 and e (1− C) −d (A+B+D) > 0. Proof. If k,o are even and I is odd positive integers, then Xn = Xn-k = Xn-o and Xn+1 and Xn+1 = Xn-1. It follows from Eq. (1) that and P= (A+B+ D) Q+ CP Q= (A+B+D) P+CQ Consequently, we get - b CP+Q=(1-0)- #1 bQ (e P- dQ)' bP (e Q- dp)* [e (1 − C) −d (A+B+D)]' (40) (41) (42)
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 12.If k,o are even and 1 is odd positive integers,
then Eq. (1) has prime period two solution if the condition
(3e- d) (1– C)< (e+d)(A+B+D),
(39)
is valid, provided C<1 and e (1- C)-d(A+B+D) >
0.
Proof.If k,o are even and 1 is odd positive integers, then
Xn-1. It follows from Eq. (1)
Xn = Xn-k = Xn-o and Xn+1
that
bQ
P= (A+B+D) Q+CP
(40)
(e P- dQ)’
and
bP
Q= (A+B+D) P+CQ –
(41)
(e Q- dP)
Consequently, we get
b
P+ Q=
(42)
[e (1-C)-d (A+B+D)]´
where e (1 – C) – d (A+B+D) > 0,
eb (1–C)
PQ=
(43)
(e+ d) [(1– C) + K4] [e (1 – C )– d K4]²
(A+B+D), provided C < 1. Substituting
where K4
(42) and (43) into (28), we get the condition (39). 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%2F6d6e12a5-1c6b-4df7-8b9f-f3408318fe51%2Fzzkglpi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 12.If k,o are even and 1 is odd positive integers,
then Eq. (1) has prime period two solution if the condition
(3e- d) (1– C)< (e+d)(A+B+D),
(39)
is valid, provided C<1 and e (1- C)-d(A+B+D) >
0.
Proof.If k,o are even and 1 is odd positive integers, then
Xn-1. It follows from Eq. (1)
Xn = Xn-k = Xn-o and Xn+1
that
bQ
P= (A+B+D) Q+CP
(40)
(e P- dQ)’
and
bP
Q= (A+B+D) P+CQ –
(41)
(e Q- dP)
Consequently, we get
b
P+ Q=
(42)
[e (1-C)-d (A+B+D)]´
where e (1 – C) – d (A+B+D) > 0,
eb (1–C)
PQ=
(43)
(e+ d) [(1– C) + K4] [e (1 – C )– d K4]²
(A+B+D), provided C < 1. Substituting
where K4
(42) and (43) into (28), we get the condition (39). Thus,
the proof is now completed.O
![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%2Fcb19240c-b74e-434f-b44c-0f0733785fa6%2F6d6e12a5-1c6b-4df7-8b9f-f3408318fe51%2Fvg0n4hv_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.
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