Solve the question in the same way as a book Plot the following curves, locate any fixed points, and obtain asymp-totic expansions to their solution as k(a) Yk+1 = y% +a², a =Yk (y%+3)3y+1(c) 2y%+1 - 5yk+1Yk + 2y% = 2,real number,(b) Ук+1 —(d) ук+1ayk+1Ykta »(е) Ук+15y -6yk+26y -8yk+3'(f) 5yk+15y+6yk+19 Example DThe equationYk+1 =1/2yk +2-3/2yk(2.193)76Difference Equationscorresponds to a hyperbola in the (yk, Yk+1) plane having asymptotes Yk = 0and yk+1 =slopes at (1, 1) and (3, 3) are, respectively, 2 and 2/3. Therefore, Yk = 1 is anunstable fixed point, while yk = 3 is a stable fixed point.A geometric analysis indicates that arbitrary starting points lead finallyto the stable fixed point. See Figure 2.10.The first approximation to the solution of equation (2.193), near yk = 3,is given by the expression1/2yk + 2. There are two fixed points, yk = 1 and yk = 3. TheYk = 3+ A(2/3)*,(2.194)where A is an arbitrary constant. Therefore, if we lett = A(2/3)*, 2(t) = Yk,(2.195)then equation (2.193) becomes2t2:(0): ()2(t)? + 4z(t) – 3.(2.196)=We assume a solution of the formz(t) = 3+t+ A2t? + Azt +....(2.197)Substitution of this expression into equation (2.197) gives, respectively, forthe left- and right-hand sides2z(t)z2t= 18 + 10t + 2/3(13A2 + 2)t+ 2/%(35A3 + 10A2)t² + ...,(2.198)andz(t)? + 4z(t) – 3 = 18 + 10t + (10A2 + 1)t?+ (10A3 + 2A2)t³ + ....(2.199)Equating the coefficients of corresponding powers of t gives2/3(13A2 + 2) = 10A2 + 1,(2.200)2/9(35A3 + 10A2) = 10A3 + 2A2,andA2 = 1/4, A3 =140.(2.201)Therefore,z(t) = 3+t+at2 + 1/40t + ... ,andYk = 3+ A(2/3)* +14A? (2/3)2k + 1/40A° (2/3)3k + ....(2.202)

Name: Solve the question in the same way as a book Plot the following curves
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Description: Solve the question in the same way as a book Plot the following curves, locate any fixed points, and obtain asymp-totic expansions to their solution as k(a) Yk+1 = y% +a², a =Yk (y%+3)3y+1(c) 2y%+1 - 5yk+1Yk + 2y% = 2,real number,(b) Ук+1 —(d) ук+1ay

As you can see in the figure attached there is only one fixed point and when k goes to infinity the…

Answered: 5y-6yk+2 6y-8yk+3) (e) Yk+1

Math

Advanced Math

5y-6yk+2 6y-8yk+3) (e) Yk+1

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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Concept explainers

Unitary Method

The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.

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Profit and Loss

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Units and Measurements

Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experiments to set up the devices.

Topic Video

Question

Solve the question in the same way as a book

Plot the following curves, locate any fixed points, and obtain asymp-
totic expansions to their solution as k
(a) Yk+1 = y% +a², a =
Yk (y%+3)
3y+1
(c) 2y%+1 - 5yk+1Yk + 2y% = 2,
real number,
(b) Ук+1 —
(d) ук+1
ayk+1
Ykta »
(е) Ук+1
5y -6yk+2
6y -8yk+3'
(f) 5yk+1
5y+6yk+19

Example D
The equation
Yk+1 =
1/2yk +2-3/2yk
(2.193)
76
Difference Equations
corresponds to a hyperbola in the (yk, Yk+1) plane having asymptotes Yk = 0
and yk+1 =
slopes at (1, 1) and (3, 3) are, respectively, 2 and 2/3. Therefore, Yk = 1 is an
unstable fixed point, while yk = 3 is a stable fixed point.
A geometric analysis indicates that arbitrary starting points lead finally
to the stable fixed point. See Figure 2.10.
The first approximation to the solution of equation (2.193), near yk = 3,
is given by the expression
1/2yk + 2. There are two fixed points, yk = 1 and yk = 3. The
Yk = 3+ A(2/3)*,
(2.194)
where A is an arbitrary constant. Therefore, if we let
t = A(2/3)*, 2(t) = Yk,
(2.195)
then equation (2.193) becomes
2t
2:(0): ()
2(t)? + 4z(t) – 3.
(2.196)
=
We assume a solution of the form
z(t) = 3+t+ A2t? + Azt +....
(2.197)
Substitution of this expression into equation (2.197) gives, respectively, for
the left- and right-hand sides
2z(t)z
2t
= 18 + 10t + 2/3(13A2 + 2)t
+ 2/%(35A3 + 10A2)t² + ...,
(2.198)
and
z(t)? + 4z(t) – 3 = 18 + 10t + (10A2 + 1)t?
+ (10A3 + 2A2)t³ + ....
(2.199)
Equating the coefficients of corresponding powers of t gives
2/3(13A2 + 2) = 10A2 + 1,
(2.200)
2/9(35A3 + 10A2) = 10A3 + 2A2,
and
A2 = 1/4, A3 =
140.
(2.201)
Therefore,
z(t) = 3+t+at2 + 1/40t + ... ,
and
Yk = 3+ A(2/3)* +14A? (2/3)2k + 1/40A° (2/3)3k + ....
(2.202)

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