Assume that P and Q are two positive distinct real roots of the quadratic equation (²₂. - - (P+Q)t+PQ=0. Thus, we deduce that C (P+Q)² > 4PQ. (27) (28)
Assume that P and Q are two positive distinct real roots of the quadratic equation (²₂. - - (P+Q)t+PQ=0. Thus, we deduce that C (P+Q)² > 4PQ. (27) (28)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Show me the determine blue and the inf is here
![Pr
Theorem 9.If k is even and 1, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
the
(1- (C+D)) (3e- d) < (e+d) (A+ B),
(20)
is
valid,
provided
(C+D)
< 1
and
e (1- (C+D)) – d (A+B) > 0.
an
Proof.If k is even and 1,o are odd positive integers, then
Xn = Xn-k and xn+1 = Xn-= Xp-g. It follows from Eg. (1)
that
Co
bQ
P= (A+B) Q+ (C+D)P –
(21)
(еР — d@)"
and
wh
bP
Q= (A+B)P+(C+D) Q –
(22)
(e Q– dP)
Consequently, we get
wł
еР- dPО %3D е (А+ B) РО— d (А+ В) @+e(С+D)P
Su
-(C+D) dPQ– bQ,
(23)
(2
and
TI
e Q – dPQ = e (A+B) PQ– d (A+B) P + e(C+D) Q
- (C+D) dPQ– bP.
in
CO
(24)
By subtracting (24) from (23), we get
b
is
(25)
[e (1– (C+D)) – d (A+B)]'
P+Q=
0.
Pr
where e (1- (C+D)) – d (A+ B) > 0. By adding (23)
and (24), we obtain
tha
eb (1- (C+D))
(e+d) [K1+ (A+B)] [e K1 – d (A+B)]²'
PQ=
(26)
(1– (C+D)), provided (C+D) < 1.
where K1
Assume that P and Q are two positive distinct real roots
of the quadratic equation
-
2-(P+Q)t+ PQ=0.
(27)
|
Thus, we deduce that
(P+Q)² > 4PQ.
(28)
Substituting (25) and (26) into (28), we get the condition
(20). Thus, the proof is now completed.O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F5a18327d-1117-42bd-bd28-e8b289ed3fae%2Fjwlsey9_processed.png&w=3840&q=75)
Transcribed Image Text:Pr
Theorem 9.If k is even and 1, o are odd positive integers,
then Eq. (1) has prime period two solution if the condition
the
(1- (C+D)) (3e- d) < (e+d) (A+ B),
(20)
is
valid,
provided
(C+D)
< 1
and
e (1- (C+D)) – d (A+B) > 0.
an
Proof.If k is even and 1,o are odd positive integers, then
Xn = Xn-k and xn+1 = Xn-= Xp-g. It follows from Eg. (1)
that
Co
bQ
P= (A+B) Q+ (C+D)P –
(21)
(еР — d@)"
and
wh
bP
Q= (A+B)P+(C+D) Q –
(22)
(e Q– dP)
Consequently, we get
wł
еР- dPО %3D е (А+ B) РО— d (А+ В) @+e(С+D)P
Su
-(C+D) dPQ– bQ,
(23)
(2
and
TI
e Q – dPQ = e (A+B) PQ– d (A+B) P + e(C+D) Q
- (C+D) dPQ– bP.
in
CO
(24)
By subtracting (24) from (23), we get
b
is
(25)
[e (1– (C+D)) – d (A+B)]'
P+Q=
0.
Pr
where e (1- (C+D)) – d (A+ B) > 0. By adding (23)
and (24), we obtain
tha
eb (1- (C+D))
(e+d) [K1+ (A+B)] [e K1 – d (A+B)]²'
PQ=
(26)
(1– (C+D)), provided (C+D) < 1.
where K1
Assume that P and Q are two positive distinct real roots
of the quadratic equation
-
2-(P+Q)t+ PQ=0.
(27)
|
Thus, we deduce that
(P+Q)² > 4PQ.
(28)
Substituting (25) and (26) into (28), we get the condition
(20). Thus, the proof is now completed.O
![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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ecaae78-467a-4f8b-9627-a81f9986c070%2F5a18327d-1117-42bd-bd28-e8b289ed3fae%2Fsw75ji_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
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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