How to deduce this equation from Equation 1 Explain to me the method. Show me the steps of determine blue and equation 1 is second pic Theorem 11.If 1,0are even and k is odd positiveintegers, then Eq.(1) has prime period two solution if thecondition(А+С+D) (Зе— d) < (е+ d) (1—В),(34)is valid, provided B< 1 and d (1 – B) – e (A+C+D) >0.Proof.If 1,0 are even and k is odd positive integers, thenXn = Xn-1= Xn-o and xn+1 = Xp-k. It follows from Eq.(1)thatbPP=(A+C+D)Q+BP(35)(e Q– dP)’andQ= (A+C+ D) P+BQbQ(еР — d0)"(36)Consequently, we getbP+Q=(37)[d (1– B) – e (A+C+D)]'where d (1- B) - e (A+C+D) > 0,eb (A+C+D)(e+d) [(1– B) + K3] [d (1– B) – e K3]?PQ=(38)where K3(37) and (38) into (28), we get the condition (34). Thus,the proof is now completed.O(A+C+D), provided B< 1. Substituting Thus, we deduce that(P+ Q)² > 4PQ.(28)The objective of this article is to investigate somequalitative behavior of the solutions of the nonlineardifference equationbxn-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), whilek, 1 and o are positive integers. The initial conditionsX-6…, X_1,..., X_k, ..., X_1, Xo are arbitrary positive realnumbers such that k <1< o. Note that the special casesof 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] whenB= C= D=0, and k= 0, b is replaced by[33] when B = C = D = 0, 1 = 0 and in [32] whenA = C= D=0, 1= 0, b is replaced by – b.(1)b and in%3D

Answered: Theorem 11.If 1,0 are even and k is odd…

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

How to deduce this equation from Equation 1 Explain to me the method. Show me the steps of determine blue and equation 1 is second pic

Theorem 11.If 1,0
are even and k is odd positive
integers, then Eq.(1) has prime period two solution if the
condition
(А+С+D) (Зе— d) < (е+ d) (1—В),
(34)
is valid, provided B< 1 and d (1 – B) – e (A+C+D) >
0.
Proof.If 1,0 are even and k is odd positive integers, then
Xn = Xn-1= Xn-o and xn+1 = Xp-k. It follows from Eq.(1)
that
bP
P=(A+C+D)Q+BP
(35)
(e Q– dP)’
and
Q= (A+C+ D) P+BQ
bQ
(еР — d0)"
(36)
Consequently, we get
b
P+Q=
(37)
[d (1– B) – e (A+C+D)]'
where d (1- B) - e (A+C+D) > 0,
eb (A+C+D)
(e+d) [(1– B) + K3] [d (1– B) – e K3]?
PQ=
(38)
where K3
(37) and (38) into (28), we get the condition (34). Thus,
the proof is now completed.O
(A+C+D), provided B< 1. Substituting

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

Expert Solution

Step 1

Given nonlinear difference equation is

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

Given that $I, σ$ are even $k$ is odd positive integer, then (1) has prime period solution.

Let the solutions are $P, Q$ .

If $x_{n + 1} = P$ , then $x_{n} = x_{n - I} = x_{n - σ} = Q, x_{n - k} = P$

So equation (1) will be

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

If $x_{n + 1} = Q$ , then $x_{n} = x_{n - I} = x_{n - σ} = P, x_{n - k} = Q$

Then from (1)

$Q = A P + B Q + C P + D P - \frac{b Q}{e P - d Q} \Rightarrow Q = (A + C + D) P + B Q - \frac{b Q}{e P - d Q} \Rightarrow e P Q - d Q^{2} = e (A + C + D) P^{2} - d (A + C + D) P Q + e B P Q - d B Q^{2} - b Q (3)$

subtracting (2)-(3)

$- d (P^{2} - Q^{2}) = - e (A + C + D) (P^{2} - Q^{2}) - d B (P^{2} - Q^{2}) - b (P - Q) \Rightarrow d (P + Q) = e (A + C + D) (P + Q) + d B (P + Q) + b \Rightarrow (P + Q) (d - e (A + C + D) - d B) = b \Rightarrow P + Q = \frac{b}{[d (1 - B) - e (A + C + D)]} (4)$

adding (2)+(3)

$2 e P Q - d (P^{2} + Q^{2}) = e (A + C + D) (P^{2} + Q^{2}) - 2 d (A + C + D) P Q + 2 e B P Q - d B (P^{2} + Q^{2}) - b (P + Q) \Rightarrow (P^{2} + Q^{2}) [- d - e (A + C + D) + d B] = 2 P Q [- d (A + C + D) + e B - e] - b (P + Q) \Rightarrow (P^{2} + Q^{2}) [- d (1 - B) - e (A + C + D)] = 2 P Q [- d (A + C + D) - e (1 - B)] - b (P + Q) \Rightarrow P^{2} + Q^{2} = \frac{2 P Q [d (A + C + D) + e (1 - B)] + b (P + Q)}{[d (1 - B) + e (A + C + D)]} (5)$