Ck De+1 = Ck-1De + kCrDe, e rewritten as De+1 De Ck-1 + kCk a, Ck

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section5.6: Exponential And Logarithmic Equations
Problem 64E
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Explain the determine green

5.3.4 Example D
Lagrange's method cannot be applied to the equation
2(k, l+ 1) = 2(k – 1, l) + kz(k, l),
(5.97)
since one of the coefficients depends on k. However, if we let z(k, l) = CkDe,
then
CkDe+1= Ck-1De + kChDe,
(5.98)
which can be rewritten as
De+1
De
Ck-1+ kCk
a,
(5.99)
where a is an arbitrary constant. Therefore, De and C 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?
2(k, l) = (-)* L T|k – (a – 1)]’
(5.102)
where o is an arbitrary function of a.
Transcribed Image Text:5.3.4 Example D Lagrange's method cannot be applied to the equation 2(k, l+ 1) = 2(k – 1, l) + kz(k, l), (5.97) since one of the coefficients depends on k. However, if we let z(k, l) = CkDe, then CkDe+1= Ck-1De + kChDe, (5.98) which can be rewritten as De+1 De Ck-1+ kCk a, (5.99) where a is an arbitrary constant. Therefore, De and C 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? 2(k, l) = (-)* L T|k – (a – 1)]’ (5.102) where o is an arbitrary function of a.
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