2.2.2 Example B Now consider the inhomogeneous equation Yk+1 - Byk = a, (2.12) 3 and gk = a. where a and 3 are constants. For this case, we have p Therefore, k-1 k-1 k-1 I, ПР Ii Pr (2.13) Using the fact that k-1 r – pk Σ (2.14) 1-r i=1 gives k-1 1- B-k+1 EB-i (2.15) B -1 Therefore, for B # 1, the general solution is a Yk = C3k (2.16) В —1 where C is an arbitrary constant. When 3 = 1, we have Yk+1 – Yk = , (2.17) with Pk = 1 and q = a. Now k-1 k-1 II Pk = II 1= 1, (2.18) i=1 i=1 and k-1 k-1 = a (1) = a(k - 1). (2.19) li Pr | r=1 i=1 IWI

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
icon
Related questions
icon
Concept explainers
Question
Explain this
2.2.2
Example B
Now consider the inhomogeneous equation
Yk+1 – Byk = a,
(2.12)
where a and B are constants. For this case, we have pk
Therefore,
3 and gk = a.
71
k-1
k-1
k-1
qi
Σ.
(2.13)
Pr
i=1
r=1
i=1
i=1
Using the fact that
k-1
pk
(2.14)
1- r
i=1
gives
k-1
1– 3-k+1
В — 1
(2.15)
Therefore, for 3+ 1, the general solution is
Yk =
(2.16)
В — 1
where C is an arbitrary constant.
When B = 1, we have
Yk+1 – Yk = a,
(2.17)
with pr = 1 and q = a. Now
k-1
k-1
SII Pk = II 1 = 1,
(2.18)
i=1
i=1
and
k-1
/k-1
k-1
> (1) = a(k – 1).
(2.19)
i=1
r=1
i=1
48
Difference Equations
Therefore, the general solution, for 3 = 1, is
Yk = A + a(k – 1) = C + ak,
(2.20)
where A and C = A – a are arbitrary constants.
IWI
Transcribed Image Text:2.2.2 Example B Now consider the inhomogeneous equation Yk+1 – Byk = a, (2.12) where a and B are constants. For this case, we have pk Therefore, 3 and gk = a. 71 k-1 k-1 k-1 qi Σ. (2.13) Pr i=1 r=1 i=1 i=1 Using the fact that k-1 pk (2.14) 1- r i=1 gives k-1 1– 3-k+1 В — 1 (2.15) Therefore, for 3+ 1, the general solution is Yk = (2.16) В — 1 where C is an arbitrary constant. When B = 1, we have Yk+1 – Yk = a, (2.17) with pr = 1 and q = a. Now k-1 k-1 SII Pk = II 1 = 1, (2.18) i=1 i=1 and k-1 /k-1 k-1 > (1) = a(k – 1). (2.19) i=1 r=1 i=1 48 Difference Equations Therefore, the general solution, for 3 = 1, is Yk = A + a(k – 1) = C + ak, (2.20) where A and C = A – a are arbitrary constants. IWI
Expert Solution
steps

Step by step

Solved in 2 steps

Blurred answer
Knowledge Booster
Points, Lines and Planes
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,