2.2.6 Example F The equation (k + 1)yk+1 – kYk = k (2.35) can be written as A(kyk) = k. (2.36) Therefore, kyk = A-1(k) = 2k(k – 1) + A, (2.37) %3D where A is an arbitrary constant. Dividing by k gives Yk = A/k+ /½(k – 1). (2.38) Equation (2.35) can also be written as k k Yk+1 Yk k+1 k +1 (2.39) Therefore, k Pk = qk = (2.40) k +1 50 Difference Equations and k-1 IIP: 1 Pi = (2.41) i=1 k-1 k-1 i i ) -E Pr (i+1) i +1 i=1 i=1 r=1 (2.42) k-1 k(k – 1) i=1 Substitution of equations (2.41) and (2.42) into equation (2.9) gives equation (2.38).
2.2.6 Example F The equation (k + 1)yk+1 – kYk = k (2.35) can be written as A(kyk) = k. (2.36) Therefore, kyk = A-1(k) = 2k(k – 1) + A, (2.37) %3D where A is an arbitrary constant. Dividing by k gives Yk = A/k+ /½(k – 1). (2.38) Equation (2.35) can also be written as k k Yk+1 Yk k+1 k +1 (2.39) Therefore, k Pk = qk = (2.40) k +1 50 Difference Equations and k-1 IIP: 1 Pi = (2.41) i=1 k-1 k-1 i i ) -E Pr (i+1) i +1 i=1 i=1 r=1 (2.42) k-1 k(k – 1) i=1 Substitution of equations (2.41) and (2.42) into equation (2.9) gives equation (2.38).
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
Topic Video
Question
Explain this Q the last one
Transcribed Image Text:2.2.6 Example F
73
The equation
(k +1)yk+1 – kYk = k
(2.35)
can be written as
A(kyk) = k.
(2.36)
Therefore,
kyk = A-1(k) = ½k(k – 1) + A,
(2.37)
where A is an arbitrary constant. Dividing by k gives
= A/k+ ½(k – 1).
(2.38)
Equation (2.35) can also be written as
k
Yk
k +1
k
Yk+1
(2.39)
k +1
Therefore,
k
Pk = qk
(2.40)
%3D
k +1
50
Difference Equations
and
k-1
1
II P: = T
(2.41)
k
i=1
k-1
i
k-1
Σ
(i + 1)
i +1
i=1
Ii
Pr
r=1
(2.42)
k-1
k(k – 1)
-Σ
2
i=1
Substitution of equations (2.41) and (2.42) into equation (2.9) gives equation
(2.38).
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Knowledge Booster
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 Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
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…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
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…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
