e b (A+C+D) (38) PQ= (e+d) [(1– B) +K3] [d (1 – B) – e K3]²
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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Show me the steps of determine red and the inf is here
![Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
difference equation
bxn-k
Xn+1 = Axn+ Bxn-k+Cxr-1+Dxn-o+
[dxn-k- exp-1]
(1)
n = 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-o,..., 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%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F3dbd883b-08c7-4197-9dde-d28dbe225b37%2Fc092aum_processed.png&w=3840&q=75)
Transcribed Image Text:Thus, we deduce that
(P+ Q)² > 4PQ.
(28)
difference equation
bxn-k
Xn+1 = Axn+ Bxn-k+Cxr-1+Dxn-o+
[dxn-k- exp-1]
(1)
n = 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-o,..., 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.
![Theorem 11.If 1,0
are even and k is odd positive
integers, then Eq. (1) has prime period two solution if the
condition
(A+C+D) (3e- d) < (e+d) (1– B),
(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 = Xp-1= Xr-o and xŋ+1 = Xn-k• It follows from Eq.(1)
that
bP
P=(A+C+D)Q+BP
(35)
(e Q- dP)'
and
bQ
Q= (A+C+D) P+BQ –
(36)
(eP
.(OP
(еР — dQ)
Consequently, we get
b
P+Q=
(37)
[d (1— В) — е
- e (A+C+D)]’
where d (1- B) – e (A+C+D) > 0,
e b (A+C+D)
PQ=
(38)
%3D
(e+d) [(1– B) + K3] [d (1 – B) – e K3]²
-
where K3 = (A+C+D), provided B< 1. Substituting
(37) and (38) into (28), we get the condition (34). Thus,
the proof is now completed.O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F3dbd883b-08c7-4197-9dde-d28dbe225b37%2Fdh21a8g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 11.If 1,0
are even and k is odd positive
integers, then Eq. (1) has prime period two solution if the
condition
(A+C+D) (3e- d) < (e+d) (1– B),
(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 = Xp-1= Xr-o and xŋ+1 = Xn-k• It follows from Eq.(1)
that
bP
P=(A+C+D)Q+BP
(35)
(e Q- dP)'
and
bQ
Q= (A+C+D) P+BQ –
(36)
(eP
.(OP
(еР — dQ)
Consequently, we get
b
P+Q=
(37)
[d (1— В) — е
- e (A+C+D)]’
where d (1- B) – e (A+C+D) > 0,
e b (A+C+D)
PQ=
(38)
%3D
(e+d) [(1– B) + K3] [d (1 – B) – e K3]²
-
where K3 = (A+C+D), provided B< 1. Substituting
(37) and (38) into (28), we get the condition (34). Thus,
the proof is now completed.O
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