Explain the determine blue

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Explain the determine blue

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125
Theorem 13.Let {xn} be a solution of Eq.(1). Then the
following
statements
are
true:
36)
b < d
(i)
N>0, the intial conditions
Suppose
and
for
some
XN-1+1,..., XN–k+1,……, XN-1, XN €
1
37)
are valid, then for b+ e and d² # be, we have the
inequality
(A+B+C+D) +
b
< Xn < (A+B+C+D)+
& – be)
(b-e) »
(44)
38)
for
all
n
N.
ng
us,
b > d
Suppose
N>0, the intial conditions
(ii)
and
for
some
ers,
XN-1+1, ..., хN-k+1,..., Xм-1, Xу €
1.
UO
39)
are valid, then for b+ e and d + be, we have the
inequality
>
b
(A+B+C+D)+o S Xn < (A+B+C+D)+
(df – be) '
(45)
(b-e)
en
for all n> N.
(1)
Proof.First of all, if for some N>0, 2< xN < 1 and b#e,
we have
10)
XN+1 = AxN+ BxN-k+CxN–i+DxN-o+
bxN-k
dxn-k– exN-1
bxN-k
< A+B+C+D+
(46)
dxN-k- exN-1
1)
But, it is easy to see that dxN-k– exN-12 b– e, then for
b#e, we get
b
XN+1 < A+B+C+D+
(47)
b-e'
12)
Similarly, we can show that
(w-1 2(A+ B+C+ D)-
b
bxN-k
XN+1 >
d
(48)
dxN-k — еxN-1
13)
&– be
then for d +
But, one can see that dxN-k– eXN–1<
be, we get
ng
us,
b
(A+B+C+D)+
XN+1 >
(49)
d – be
From (47) and (49) we deduce for all n> N that the
inequality (44) is valid. Hence, the proof of part (i) is
completed.
Similarly, if 1 < XN < , then we can prove part (ii)
which is omitted here for convenience. Thus, the proof is
now completed.O
che
A 201 4 NSP
AL
Transcribed Image Text:ishing.com/Journals.asp 125 Theorem 13.Let {xn} be a solution of Eq.(1). Then the following statements are true: 36) b < d (i) N>0, the intial conditions Suppose and for some XN-1+1,..., XN–k+1,……, XN-1, XN € 1 37) are valid, then for b+ e and d² # be, we have the inequality (A+B+C+D) + b < Xn < (A+B+C+D)+ & – be) (b-e) » (44) 38) for all n N. ng us, b > d Suppose N>0, the intial conditions (ii) and for some ers, XN-1+1, ..., хN-k+1,..., Xм-1, Xу € 1. UO 39) are valid, then for b+ e and d + be, we have the inequality > b (A+B+C+D)+o S Xn < (A+B+C+D)+ (df – be) ' (45) (b-e) en for all n> N. (1) Proof.First of all, if for some N>0, 2< xN < 1 and b#e, we have 10) XN+1 = AxN+ BxN-k+CxN–i+DxN-o+ bxN-k dxn-k– exN-1 bxN-k < A+B+C+D+ (46) dxN-k- exN-1 1) But, it is easy to see that dxN-k– exN-12 b– e, then for b#e, we get b XN+1 < A+B+C+D+ (47) b-e' 12) Similarly, we can show that (w-1 2(A+ B+C+ D)- b bxN-k XN+1 > d (48) dxN-k — еxN-1 13) &– be then for d + But, one can see that dxN-k– eXN–1< be, we get ng us, b (A+B+C+D)+ XN+1 > (49) d – be From (47) and (49) we deduce for all n> N that the inequality (44) is valid. Hence, the proof of part (i) is completed. Similarly, if 1 < XN < , then we can prove part (ii) which is omitted here for convenience. Thus, the proof is now completed.O che A 201 4 NSP AL
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn– k
X+1 = Axn+ Bxp–k+CXp-1+Dxp-o+
[dxn-k– ex-1
(1)
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-g,..., 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:The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxn– k X+1 = Axn+ Bxp–k+CXp-1+Dxp-o+ [dxn-k– ex-1 (1) 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-g,..., 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.
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