show me the steps of determine blue and the inf is here Theorem 13.Let {Xn} be a solution of Eq. (1). Then thefollowingstatementsaretrue:(1)N>0, the intial conditionsSupposeb <andforsomeXN-1+1, ..., XN-k+1,·.., XN-1, XN Eare valid, then for b + e and d + be, we have theinequalityb(b-e)(44)$(A+B+C+D)+< Xn < (A+B+C+D)+- be)forallnN.(ii)N>0, the intial conditionsSupposeb >dandforsomeXN-1+1,..,XN-k+1,..., XN-1, XN E| 1,valid, then for be and d + be,we have theinequality(A+B+C+D)+ < Xp(A+B+C+D) + Tb(b-e)(d-be)(45)for all n> N.Proof.First of all, if for some N>0, 2< XN <1 and b#e,we havebxN-kXN+1 = AXN+ BXN–k+ CxN–1+DxN-o +dxN-k- exN-1bxN-k N that theinequality (44) is valid. Hence, the proof of part (i) iscompleted.Similarly, if 1 < XN <, then we can prove part (ii)which is omitted here for convenience. Thus, the proof isnow completed.O The main focus of this article is to discuss some qualitative behavior ofthe solutions of the nonlinear difference equationa1Ym-1+a2Ym-2+ a3Ym-3+ a4Ym–4+ a5Ym-5Ут+1 — Аутtт 3 0, 1, 2, ...,B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5(1.1)where the coefficients A, a;, B; E (0, 0), i = 1, ..., 5, while the initial condi-tions y-5,y-4,Y–3,Y–2, Y–1, yo are arbitrary positive real numbers. Note thatthe special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 == a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special caseB4when a4 =B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in thespecial case when az = B5 = 0.

To get the inequality (48). XN+1≥bdA+B+C+D+bXN−kdXN−k−eXN−1 .......................48…

Answered: Theorem 13.Let {xn} be a solution of…

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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show me the steps of determine blue and the inf is here

Theorem 13.Let {Xn} be a solution of Eq. (1). Then the
following
statements
are
true:
(1)
N>0, the intial conditions
Suppose
b <
and
for
some
XN-1+1, ..., XN-k+1,·.., XN-1, XN E
are valid, then for b + e and d + be, we have the
inequality
b
(b-e)
(44)
$(A+B+C+D)+
< Xn < (A+B+C+D)+
- be)
for
all
n
N.
(ii)
N>0, the intial conditions
Suppose
b >
d
and
for
some
XN-1+1,..,XN-k+1,..., XN-1, XN E| 1,
valid, then for b
e and d + be,
we have the
inequality
(A+B+C+D)+ < Xp(A+B+C+D) + T
b
(b-e)
(d-be)
(45)
for all n> N.
Proof.First of all, if for some N>0, 2< XN <1 and b#e,
we have
bxN-k
XN+1 = AXN+ BXN–k+ CxN–1+DxN-o +
dxN-k- exN-1
bxN-k
<A+B+C+D+
(46)
dxN-k- exN-1
But, it is easy to see that dxN-k– exN-12 b- e, then for
b+ e, we get
XN+1 <A+B+C+D+
(47)
b-e
е
Similarly, we can show that
XN+1 2
b
(A+B+C+ D) +
bxN-k
(48)
dxN-k- exN-1
But, one can see that dxN-k- exN–1<be, then for d +
be, we get
d.
XN+1 2
(A+B+C+D) +
(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

The main focus of this article is to discuss some qualitative behavior of
the solutions of the nonlinear difference equation
a1Ym-1+a2Ym-2+ a3Ym-3+ a4Ym–4+ a5Ym-5
Ут+1 — Аутt
т 3 0, 1, 2, ...,
B1Ym-1+ B2Ym-2 + B3Ym-3 + B4Ym-4 + B3Ym-5
(1.1)
where the coefficients A, a;, B; E (0, 0), i = 1, ..., 5, while the initial condi-
tions y-5,y-4,Y–3,Y–2, Y–1, yo are arbitrary positive real numbers. Note that
the special case of Eq.(1.1) has been discussed in [4] when az = B3 = a4 =
= a5 = B5 = 0 and Eq.(1.1) has been studied in [8] in the special case
B4
when a4 =
B4 = a5 = B5 = 0 and Eq.(1.1) has been discussed in [5] in the
special case when az = B5 = 0.

Expert Solution

Step 1

To get the inequality (48).

$X_{N + 1} \geq \frac{b}{d} (A + B + C + D) + \frac{b X_{N - k}}{d X_{N - k} - e X_{N - 1}} ....................... (48)$

We have

$\begin{array}{l} x_{N + 1} = A X_{N} + B X_{N - k} + C x_{N - l} + D X_{N - σ} + \frac{b X_{N - k}}{d X_{N - k} - e X_{n - l}} ............ (*) \\ where the coefficients A, B, C, D, b, d, e \in (0, \infty) \\ and k, l, and σ are positive inegers \end{array}$

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