Explain the detemine green
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Explain the detemine green
![5.3.4 Example D
Lagrange's method cannot be applied to the equation
z(k,l+1) = z(k – 1, €) + kz(k, l),
(5.97)
since one of the coefficients depends on k. However, if we let z(k, e) = C1,De,
then
CrDe+1 = Ck-1D + kCrDt,
(5.98)
which can be rewritten as
De+1
Ck-1 + kCk
a,
(5.99)
%3D
De
where a is an arbitrary constant. Therefore, De and C satisfy the first-order
difference equations
De+1 = aDt, (a – k)Ck = Ck-1;
(5.100)
the solutions of which are, respectively,
B(-1)*
T(k – (a – 1)]'
De = Ao', Ck =
(5.101)
where A and B are arbitrary constants. Summing over a gives
$(a)a²
z(k, l) = (-)* Lrk – (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%2F25a3cfb3-619d-471e-bb5a-8c85b1316cb2%2Fd676374a-a3f8-4844-8abe-6d26eac6f8a7%2F7cg854_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, €) + kz(k, l),
(5.97)
since one of the coefficients depends on k. However, if we let z(k, e) = C1,De,
then
CrDe+1 = Ck-1D + kCrDt,
(5.98)
which can be rewritten as
De+1
Ck-1 + kCk
a,
(5.99)
%3D
De
where a is an arbitrary constant. Therefore, De and C satisfy the first-order
difference equations
De+1 = aDt, (a – k)Ck = Ck-1;
(5.100)
the solutions of which are, respectively,
B(-1)*
T(k – (a – 1)]'
De = Ao', Ck =
(5.101)
where A and B are arbitrary constants. Summing over a gives
$(a)a²
z(k, l) = (-)* Lrk – (a – 1)]’
(5.102)
where o is an arbitrary function of a.
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