B(-1)k T(k – (a – 1)]? Ξ
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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Explain the determine blue
![5.3.4 Example D
Lagrange's method cannot be applied to the equation
z(k, l+1) = z(k – 1, l) + kz(k, l),
(5.97)
since one of the coefficients depends on k. However, if we let z(k,l) = CDe,
then
Ck De+1 = Ck-1De + kC;De,
(5.98)
which can be rewritten as
De+1
Ck-1+ kCk
-a,
(5.99)
De
Ck
where a is an arbitrary constant. Therefore, De and Ck satisfy the first-order
difference equations
De+1 = aDe, (a – k)Ck = Ck-1,
(5.100)
the solutions of which are, respectively,
B(-1)*
I(k – (a – 1) ?
De = Aa', (Ck
(5.101)
where A and B are arbitrary constants. Summing over a gives
$(a)a²
z(k, l) = (-)* L |k – (a – 1)]'
(5.102)
where o is an arbitrary function of a.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9ff41fc3-4717-401c-a6f6-d444dc924011%2Fbd0b620e-4a74-4a41-b42d-009822d49360%2F0413u9a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5.3.4 Example D
Lagrange's method cannot be applied to the equation
z(k, l+1) = z(k – 1, l) + kz(k, l),
(5.97)
since one of the coefficients depends on k. However, if we let z(k,l) = CDe,
then
Ck De+1 = Ck-1De + kC;De,
(5.98)
which can be rewritten as
De+1
Ck-1+ kCk
-a,
(5.99)
De
Ck
where a is an arbitrary constant. Therefore, De and Ck satisfy the first-order
difference equations
De+1 = aDe, (a – k)Ck = Ck-1,
(5.100)
the solutions of which are, respectively,
B(-1)*
I(k – (a – 1) ?
De = Aa', (Ck
(5.101)
where A and B are arbitrary constants. Summing over a gives
$(a)a²
z(k, l) = (-)* L |k – (a – 1)]'
(5.102)
where o is an arbitrary function of a.
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