Consequently, we get b P+Q= (42) [e (1 – C)– d (A+B+D)]’ where e (1– C)-d (A+B+D) > 0, eb² (1–C) PQ= (e+d) [(1– C)+ K4] [e (1– C)- d K4] (43)

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ISBN:9780470458365
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Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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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.
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 12.If k, o are even and 1 is odd positive integers,
then Eq. (1) has prime period two solution if the condition
(Зе— d) (1— с) < (е+d) (А+B+ D),
(39)
is valid, provided C< 1 and e (1-C)-d(A+B+D) >
0.
Proof.If k, o are even and 1 is odd positive integers, then
Xn = Xn-k = Xn-o and xp+1 = Xp–1. It follows from Eq.(1)
that
bQ
P= (A+B+D) Q+CP –
(40)
(е Р- dQ)
and
bP
Q= (A+B+ D)P+CQ –
(41)
(e Q- dP)
Consequently, we get
P+Q=
(42)
[e (1- C)-d (A+B+D)]’
where e (1– C)-d (A+B+ D) > 0,
eb² (1-C)
PQ=
(43)
(e+d) [(1– C) + K4] [e (1 – C)– d K4]²
where K4 = (A+B+D), provided C< 1. Substituting
(42) and (43) into (28), we get the condition (39). Thus,
the proof is now completed.O
Transcribed Image Text:Theorem 12.If k, o are even and 1 is odd positive integers, then Eq. (1) has prime period two solution if the condition (Зе— d) (1— с) < (е+d) (А+B+ D), (39) is valid, provided C< 1 and e (1-C)-d(A+B+D) > 0. Proof.If k, o are even and 1 is odd positive integers, then Xn = Xn-k = Xn-o and xp+1 = Xp–1. It follows from Eq.(1) that bQ P= (A+B+D) Q+CP – (40) (е Р- dQ) and bP Q= (A+B+ D)P+CQ – (41) (e Q- dP) Consequently, we get P+Q= (42) [e (1- C)-d (A+B+D)]’ where e (1– C)-d (A+B+ D) > 0, eb² (1-C) PQ= (43) (e+d) [(1– C) + K4] [e (1 – C)– d K4]² where K4 = (A+B+D), provided C< 1. Substituting (42) and (43) into (28), we get the condition (39). Thus, the proof is now completed.O
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