The equilibrium point of Eq.(1) is the positive solution of the equation X= (A+B+C+D)x+ (6) (d– e)x where d + e. If [(A+B+C+D) – 1] (e – d) > 0, then the only positive equilibrium point ž of Eq. (1) is given by b (7) [(A+B+C+D) – 1] (e – d)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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How to deduce this equation from Equation  1 Explain to me the method. Show me the steps of determine green and equation 1 is second pic

2 The local stability of the solutions
The equilibrium point x of Eq.(1) is the positive solution
of the equation
X= (A+B+C+D)x+
(6)
(d- e)x'
|
where d+ e. If [(A+B+C+D) – 1] (e – d) > 0, then
the only positive equilibrium point ã of Eq.(1) is given by
X=
[(A+B+C+D) – 1] (e – d) '
(7)
Let
us
now
introduce
a
continuous
function
F: (0,0)4
→ (0,0) which is defined by
6.
bu
(du1 – euz)
(8)
F(uo, U1, U2, U3) = Auo + Buj +Cu2 +Du3 +
provided du # eu2. Consequently, we get
ƏF(x,x,X,X)
= A= Po,
On e
е (А+В+С+D) -1]
(е — d)
ƏF(x,X,X,X)
= B-
= P1,
|
Ine
(9)
ƏF(x,x,x,X)
e[(A+B+C+D) –1]
(e – d)
= C+
2ne
ƏF(x,x,X,X) – D=P3,
En e
where e+ d. Thus, the linearized equation of Eq.(1) about
x takes the form
Zn+1- Pozn- P1Zn–k- P2Zn–1- P3Zn-o = 0,
(10)
where po, P1, P2 and P3 are given by (9).
Transcribed Image Text:2 The local stability of the solutions The equilibrium point x of Eq.(1) is the positive solution of the equation X= (A+B+C+D)x+ (6) (d- e)x' | where d+ e. If [(A+B+C+D) – 1] (e – d) > 0, then the only positive equilibrium point ã of Eq.(1) is given by X= [(A+B+C+D) – 1] (e – d) ' (7) Let us now introduce a continuous function F: (0,0)4 → (0,0) which is defined by 6. bu (du1 – euz) (8) F(uo, U1, U2, U3) = Auo + Buj +Cu2 +Du3 + provided du # eu2. Consequently, we get ƏF(x,x,X,X) = A= Po, On e е (А+В+С+D) -1] (е — d) ƏF(x,X,X,X) = B- = P1, | Ine (9) ƏF(x,x,x,X) e[(A+B+C+D) –1] (e – d) = C+ 2ne ƏF(x,x,X,X) – D=P3, En e where e+ d. Thus, the linearized equation of Eq.(1) about x takes the form Zn+1- Pozn- P1Zn–k- P2Zn–1- P3Zn-o = 0, (10) where po, P1, P2 and P3 are given by (9).
The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn–k
[dxp-k– exp-1]
Xn+1 =
Axn+ Bxn–k+ Cxn-1+Dxn-o +
n= 0,1,2, ...
where the coefficients A, B, C, D, b, d, e E (0,0), while
k, 1 and o are positive integers. The initial conditions
X-6.……., X_1,..., X_k ….., X_1, Xo are arbitrary positive real
numbers such that k < 1< o. 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 and in [27] when
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.
(1)
%3D
%3D
Transcribed Image Text:The objective of this article is to investigate some qualitative behavior of the solutions of the nonlinear difference equation bxn–k [dxp-k– exp-1] Xn+1 = Axn+ Bxn–k+ Cxn-1+Dxn-o + n= 0,1,2, ... where the coefficients A, B, C, D, b, d, e E (0,0), while k, 1 and o are positive integers. The initial conditions X-6.……., X_1,..., X_k ….., X_1, Xo are arbitrary positive real numbers such that k < 1< o. 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 and in [27] when 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. (1) %3D %3D
Expert Solution
Step 1

Given:

                  xn+1=Axn+Bxn-k+Cxn-l+Dxn-σ+bxn-kdxn-k-exn-l, n=0,1,2,  ...... (1)

x~ is the equilibrium point of (1) and is the positive solution of the equation

                                                          x~=A+B+C+Dx~+bx~d-ex~ .......... (6)

To find:

                                                        x~=bA+B+C+D-1e-d .......... (7)

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