What is the difference between theory 7 and theory 10 Give me a reason for that? Eq1 in the second picture Theorem 7.If k,1 are even and o is odd positive integersand (A+B+ C) +1#D, then Eq.(1) has no prime periodtwo solution.Proof.Following the proof of Theorem 5, we deduce thatif k, 1 are even and o is odd positive integers, then x, =Xn-k = Xn-1 and X+1 = Xn-o. It follows from Eq.(1) thatbP= (A+B+ C) Q+ DP –(16)(e – d)'andbQ= (A+B+C) P+DQ –(17)(e – d)'By subtracting (17) from (16), we get(P- Q) [(A+B+ C+1) – D]= 0,Since (A+B+C+1) – D #0, then P= Q. This is acontradiction. Thus, the proof is now completed.OThus, we deduce that(P+ Q)² > 4PQ.(28)Substituting (25) and (26) into (28), we get the condition(20). Thus, the proof is now completed.OTheorem 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)isd (1– (B+D)) –(B+D)provided(A+C) > 0.valid,< 1and- eProof.If 1 is even and k, o are odd positive integers, thenIt follows from Eq.(1)Xn = Xn-1andXn+1 = Xn-k= Xn-o.thatbPP= (A+C) Q+ (B+D) P –(30)(e Q- dP)'andbQQ= (A+C)P+(B+D) Q(31)(e P- dQ)'Consequently, we getbP+ 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 'PQ=(33)(1- (B+D)), provided (B+ D) < 1.where K2Substituting (32) and (33) into (28), we get the condition(29). Thus, the proof is now completed.O The objective of this article is to investigate somequalitative behavior of the solutions of the nonlineardifference equationbxn-k[dxp-k– exn=1](1)Xn+1Axn+ Bxn–k+ Cxr-1+DXn-ot= U:0,1,2,....where the coefficients A, B, C, D, b, d, e E (0,∞), whilek, 1 and o are positive integers. The initial conditionsX-0,…., X_1,..., X_k) ..., X_1, Xo are arbitrary positive realnumbers such that k < 1< 0. Note that the special casesof Eq.(1) have been studied in [1] when B=C= D= 0,•..)

Given nonlinear difference equation is xn+1=Axn+Bn-k+Cxn-l+Dxn-σ+bxn-kdxn-k-exn-l (1)…

Answered: What is the difference between theory 7…

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

What is the difference between theory 7 and theory 10 Give me a reason for that? Eq1 in the second picture

Theorem 7.If k,1 are even and o is odd positive integers
and (A+B+ C) +1#D, then Eq.(1) has no prime period
two solution.
Proof.Following the proof of Theorem 5, we deduce that
if k, 1 are even and o is odd positive integers, then x, =
Xn-k = Xn-1 and X+1 = Xn-o. It follows from Eq.(1) that
b
P= (A+B+ C) Q+ DP –
(16)
(e – d)'
and
b
Q= (A+B+C) P+DQ –
(17)
(e – d)'
By subtracting (17) from (16), we get
(P- Q) [(A+B+ C+1) – D]= 0,
Since (A+B+C+1) – D #0, then P= Q. This is a
contradiction. Thus, the proof is now completed.O
Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
Substituting (25) and (26) into (28), we get the condition
(20). Thus, the proof is now completed.O
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
d (1– (B+D)) –
(B+D)
provided
(A+C) > 0.
valid,
< 1
and
- e
Proof.If 1 is even and k, o are odd positive integers, then
It follows from Eq.(1)
Xn = Xn-1
and
Xn+1 = Xn-k= Xn-o.
that
bP
P= (A+C) Q+ (B+D) P –
(30)
(e Q- dP)'
and
bQ
Q= (A+C)P+(B+D) Q
(31)
(e P- dQ)'
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 '
PQ=
(33)
(1- (B+D)), provided (B+ D) < 1.
where K2
Substituting (32) and (33) into (28), we get the condition
(29). 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
[dxp-k– exn=1]
(1)
Xn+1
Axn+ Bxn–k+ Cxr-1+DXn-ot
= U
:0,1,2,....
where the coefficients A, B, C, D, b, d, e E (0,∞), while
k, 1 and o are positive integers. The initial conditions
X-0,…., 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,
•..)

Expert Solution

Step 1

Given nonlinear difference equation is

$x_{n + 1} = A x_{n} + B_{n - k} + C x_{n - l} + D x_{n - σ} + \frac{b x_{n - k}}{[d x_{n - k} - e x_{n - l}]}$ (1)

Since we know that prime period of two solution of the type $. . ., P, Q, P, Q, . . .$ where $P$ and $Q$ are not same.

Solution 7:

Since

$k, l$ are even and $σ$ is odd integers, then $x_{n} = x_{n - k} = x_{n - l}$ and $x_{n + 1} = x_{n - σ}$ .

If we take $x_{n + 1} = P$ then $x_{n} = x_{n - k} = x_{n - l} = Q$ and $x_{n + 1} = x_{n - σ} = P$

So from equation (1)

$P = A Q + B Q + C Q + D P + \frac{b Q}{[d Q - e Q]} \Rightarrow P = (A + B + C) Q + D P - \frac{b}{[e - d]} (2)$

If we take $x_{n + 1} = Q$ then $x_{n} = x_{n - k} = x_{n - l} = P$ and $x_{n + 1} = x_{n - σ} = Q$

then from (1)

$Q = A P + B P + C P + D Q + \frac{b P}{[d P - e P]} \Rightarrow Q = (A + B + C) P + D Q - \frac{b}{[e - d]} (3)$

Subtracting (2)-(3)

$P - Q = (A + B + C) (Q - P) + D (P - Q) \Rightarrow (P - Q) [(1 + A + B + C) - D] = 0$

Since $[(A + B + C + 1) - D] \neq 0$

So $P = Q$ .

This is not prime period solution because $P$ and $Q$ are not unique.

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