b P+Q= [d (1 — (В+ D)) — е (А+С)] where d (1– (B+D)) – e (A+C) > 0, e b² (A+C) PQ= (e+d) [K2+(A+ C)][d K2 – e (A+C)]²" -
b P+Q= [d (1 — (В+ D)) — е (А+С)] where d (1– (B+D)) – e (A+C) > 0, e b² (A+C) PQ= (e+d) [K2+(A+ C)][d K2 – e (A+C)]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 green 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%2Fb557fbbc-5698-4469-85f9-8675f781fa29%2F4wn23ae_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 10. If 1 is even and k, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(А+ C) (Зе— d) < (е+ d) (1-(В+ D),
(29)
is
valid,
(B+ D)
and
provided
d (1 — (В+ D) — е (А+С)>0.
< 1
Proof.If 1 is even and k,
are odd positive integers, then
Xn = Xp–1 and xn+1 = Xn-k = Xn-o. It follows from Eq.(1)
that
bP
Р- (А+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 — (В+ D)) — е (А+С)]"
where d (1– (B+D)) – e (A+C) > 0,
e b² (A+C)
PQ=
(e+d) [K2+(A+ C)][d K2 – e (A+C)]²"
(33)
-
where K2
(1– (B+D)), provided (B+ D) < 1.
Substituting (32) and (33) into (28), we get the condition
(29). 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%2Fb557fbbc-5698-4469-85f9-8675f781fa29%2Fqz1w1la_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Theorem 10. If 1 is even and k, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
(А+ C) (Зе— d) < (е+ d) (1-(В+ D),
(29)
is
valid,
(B+ D)
and
provided
d (1 — (В+ D) — е (А+С)>0.
< 1
Proof.If 1 is even and k,
are odd positive integers, then
Xn = Xp–1 and xn+1 = Xn-k = Xn-o. It follows from Eq.(1)
that
bP
Р- (А+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 — (В+ D)) — е (А+С)]"
where d (1– (B+D)) – e (A+C) > 0,
e b² (A+C)
PQ=
(e+d) [K2+(A+ C)][d K2 – e (A+C)]²"
(33)
-
where K2
(1– (B+D)), provided (B+ D) < 1.
Substituting (32) and (33) into (28), we get the condition
(29). Thus, the proof is now completed.O
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