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) valid, provided (C+D) and is 1 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 bP Q= (A+B) P+(C+D) Q – (22) (e Q– dP)' Consequently, we get e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+D) P² - (C+D) dPQ– bQ, (23) and e Q – dPQ = e (A+B) PQ– d (A+ B) P +e(C+D) Q - (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 b² (1 – (C+ D)) (e+d) [K1 + (A+ B)][e K1 – d (A+B)]² PQ= (26)
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) valid, provided (C+D) and is 1 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 bP Q= (A+B) P+(C+D) Q – (22) (e Q– dP)' Consequently, we get e P – dPQ = e (A+B) PQ– d (A+B) Q + e(C+D) P² - (C+D) dPQ– bQ, (23) and e Q – dPQ = e (A+B) PQ– d (A+ B) P +e(C+D) Q - (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 b² (1 – (C+ D)) (e+d) [K1 + (A+ B)][e K1 – d (A+B)]² PQ= (26)
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 green and equation 1 is second pic
![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)
valid,
provided
(C+D)
1
and
is
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 — d@)"
and
bP
Q= (A+B) P+(C+D) Q –
(22)
(e Q– dP)'
Consequently, we get
ер - dPQ —D е (А+ В) РQ— d (А+ В) + e(С+D) Р
- (C+D) dPQ– bQ,
(23)
and
e Q – dPQ = e (A+B) PQ– d (A+B) P + e(C+D) Q
- (C+D) dPQ– bP.
(24)
By subtracting (24) from (23), we get
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))
(e+d) [K1+(A+ B)][e K1 – d (A+B)]*
PQ =
(26)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fca1a5904-11c1-4e23-ad3b-bb585ae27c7a%2F765356bc-b5a3-4da2-8d2c-8e79bd51a878%2Flz62ixa_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)
valid,
provided
(C+D)
1
and
is
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 — d@)"
and
bP
Q= (A+B) P+(C+D) Q –
(22)
(e Q– dP)'
Consequently, we get
ер - dPQ —D е (А+ В) РQ— d (А+ В) + e(С+D) Р
- (C+D) dPQ– bQ,
(23)
and
e Q – dPQ = e (A+B) PQ– d (A+B) P + e(C+D) Q
- (C+D) dPQ– bP.
(24)
By subtracting (24) from (23), we get
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))
(e+d) [K1+(A+ B)][e K1 – d (A+B)]*
PQ =
(26)
![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%2F765356bc-b5a3-4da2-8d2c-8e79bd51a878%2Fl8nz92o_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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