ƏF(x,X,X.X) = A= Po, One e [(A+B+C+D) –1] (e - d) ƏF(x,x,X,X) B- Ine e((A+B+C+D) –1] (е — d) ƏF(x,X,X,X) C+ P21 ƏF(x,x,X,X) = D=P3, %3| Ene
ƏF(x,X,X.X) = A= Po, One e [(A+B+C+D) –1] (e - d) ƏF(x,x,X,X) B- Ine e((A+B+C+D) –1] (е — d) ƏF(x,X,X,X) C+ P21 ƏF(x,x,X,X) = D=P3, %3| Ene
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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Topic Video
Question
Show me the steps of determine red and the inf is here
![The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
Xn+1 = Axn+ Bx–k+Cxn-1+ Dxn-o+
dxn-k– exŋ-
n= 0,1,2,.....
(1)
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_g,..., X_1,.., X_k, ….., X_1, X 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 and in [27] when
B=C= D=0, and k= 0, b is replaced by
[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
%3|](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Ff896ab32-b807-4938-bec7-c9d063a2b250%2Fxhexhbh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The objective of this article is to investigate some
qualitative behavior of the solutions of the nonlinear
difference equation
bxn-k
Xn+1 = Axn+ Bx–k+Cxn-1+ Dxn-o+
dxn-k– exŋ-
n= 0,1,2,.....
(1)
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_g,..., X_1,.., X_k, ….., X_1, X 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 and in [27] when
B=C= D=0, and k= 0, b is replaced by
[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
%3|
![2 The local stability of the solutions
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 x of Eq.(1) is given by
b
X=
[(A+B+C+D) – 1] (e – d)
(7)
Let
us
now
introduce
a
continuous
function
F: (0, 00)4 -
→ (0,00) which is defined by
bu
F(40, u1, U2, 13) = Auo + Bun +Cu2 + Du3 +
(du
– eu2
-
(8)
provided dui # euz. Consequently, we get
ƏF(x,X,X,X)
= A= Po,
On e
e [(A+B+C+D) –1]
(e – d)
ƏF(x,X,x,X)
-
В-
P1,
%3|
(9)
e[(A+B+C+D) –1]
(e - d)
ƏF(x,x,Xx,X)
C+
P2,
ƏF(x,x,x,X)
диз
=D3DP3,
where et d. Thus, the linearized equation of Eq.(1) about
x takes the form
Zn+1 - Pozn- PıZn-k- P2Zn-1- P3Zn-o = 0,
(10)
%3D
where po, P1, P2 and p3 are given by (9).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2Ff896ab32-b807-4938-bec7-c9d063a2b250%2Ff3o6qlk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2 The local stability of the solutions
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 x of Eq.(1) is given by
b
X=
[(A+B+C+D) – 1] (e – d)
(7)
Let
us
now
introduce
a
continuous
function
F: (0, 00)4 -
→ (0,00) which is defined by
bu
F(40, u1, U2, 13) = Auo + Bun +Cu2 + Du3 +
(du
– eu2
-
(8)
provided dui # euz. Consequently, we get
ƏF(x,X,X,X)
= A= Po,
On e
e [(A+B+C+D) –1]
(e – d)
ƏF(x,X,x,X)
-
В-
P1,
%3|
(9)
e[(A+B+C+D) –1]
(e - d)
ƏF(x,x,Xx,X)
C+
P2,
ƏF(x,x,x,X)
диз
=D3DP3,
where et d. Thus, the linearized equation of Eq.(1) about
x takes the form
Zn+1 - Pozn- PıZn-k- P2Zn-1- P3Zn-o = 0,
(10)
%3D
where po, P1, P2 and p3 are given by (9).
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