e& – dPQ = e (A+B) PQ– d (A+ B) Ỷ + e(C+D) & - (C+D) dPQ– bP. - (24) -
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
Show me the steps of determine green and the inf is here
![The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
Xn+1 = Axn+ Bx–k+Cxn-1+ Dxn-o+
dxn-k– exŋ-
n= 0,1,2,.....
(1)
where the coefficients A, B, C, D, b, d, e E (0,0), while
k, 1 and o are positive integers. The initial conditions
X_g,..., X_1,.., X_k, ….., X_1, X 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
[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
%3|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Ff8d4c06e-39c2-43b2-8e6f-202c82f41de9%2Foru0tff_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
Xn+1 = Axn+ Bx–k+Cxn-1+ Dxn-o+
dxn-k– exŋ-
n= 0,1,2,.....
(1)
where the coefficients A, B, C, D, b, d, e E (0,0), while
k, 1 and o are positive integers. The initial conditions
X_g,..., X_1,.., X_k, ….., X_1, X 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
[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
%3|
![Theorem 9. If k is even and 1,0 are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(1 – (C+D)) (3e– d) < (e+d) (A+B),
(20)
is
valid,
provided
(C+D)
1
and
e (1 – (C+D)) – d (A+B) >0.
Proof.If k is even and 1, o are odd positive integers, then
Xn = Xn-k and Xn+1 = Xn–1 = Xn-o. It follows from Eq.(1)
that
bQ
P=(A+B) Q+(C+D) P
(21)
(еР — dQ)
and
БР
Q= (A+B) P+(C+D) Q –
(22)
(е Q— dP)
Consequently, we get
e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+ D) P
- (C+D) dPQ- bQ,
|
(23)
and
e – dPQ = e (A+B) PQ – d (A+ B) P + e(C+D) &
- (C+D) dPQ, bP.
(24)
By subtracting (24) from (23), we get
b
P+Q=
(25)
[e (1 – (C+D)) –d (A+B)]'
where e (1– (C+D)) – d (A+B) > 0. By adding (23)
and (24), we obtain
e B (1 – (C+D))
PQ=
(e+d)[K1+(A+ B)] [e K1 – d (A+ B)]²'
(26)
|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Ff8d4c06e-39c2-43b2-8e6f-202c82f41de9%2Fcnogkf2_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 9. If k is even and 1,0 are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(1 – (C+D)) (3e– d) < (e+d) (A+B),
(20)
is
valid,
provided
(C+D)
1
and
e (1 – (C+D)) – d (A+B) >0.
Proof.If k is even and 1, o are odd positive integers, then
Xn = Xn-k and Xn+1 = Xn–1 = Xn-o. It follows from Eq.(1)
that
bQ
P=(A+B) Q+(C+D) P
(21)
(еР — dQ)
and
БР
Q= (A+B) P+(C+D) Q –
(22)
(е Q— dP)
Consequently, we get
e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+ D) P
- (C+D) dPQ- bQ,
|
(23)
and
e – dPQ = e (A+B) PQ – d (A+ B) P + e(C+D) &
- (C+D) dPQ, bP.
(24)
By subtracting (24) from (23), we get
b
P+Q=
(25)
[e (1 – (C+D)) –d (A+B)]'
where e (1– (C+D)) – d (A+B) > 0. By adding (23)
and (24), we obtain
e B (1 – (C+D))
PQ=
(e+d)[K1+(A+ B)] [e K1 – d (A+ B)]²'
(26)
|
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