Another way to derive this same result is to start with Ck B, Ck-1 or Ck = Aßk, (8.20) %3D in equation (8.9). Substituting this function in equation (8.9) and simplifying gives CgDe+1 – BDe + (p3²)De-1 = 0, (8.21) which has the characteristic equation qr² – Br + pB? = 0. (8.22) Therefore, pB r2 = (8.23) ri = B, and De = A3(B)B² + A4 (B) ( 2 ) (8.24) Since z(k, l, B) = CkDe, we have z(k, l, B) = Ã3(8)Bktl + Ã4(8) ( 2 (8.25) where [A3(B), A4(B), A3(B), A4(B)] are arbitrary functions of B. If we now sum/integrate over B, then (2) z(k, l) = g1(k + l) + h1(k + l), (8.26) ||

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 red

does not work. To show this, consider the equation
z(k, l) = pz(k+1, l – 1) + qz(k – 1,l+1),
(8.8)
where p+q = 1. From z(k, e) = CxDe, it follows that
(은) (금)
Ck+1
Ck-1
De+1
De-1,
Ck
De
(8.9)
= p
Ck-1
De-1
UIOIIS OI u.
we now SUI/ Itegiate ovei u, te 101OWII general SOTUTIOI
п
follows
k
z(k, l) = g(k + l) +
9 h(k + e),
(8.19)
which is the same as Eq. (5.95).
Transcribed Image Text:does not work. To show this, consider the equation z(k, l) = pz(k+1, l – 1) + qz(k – 1,l+1), (8.8) where p+q = 1. From z(k, e) = CxDe, it follows that (은) (금) Ck+1 Ck-1 De+1 De-1, Ck De (8.9) = p Ck-1 De-1 UIOIIS OI u. we now SUI/ Itegiate ovei u, te 101OWII general SOTUTIOI п follows k z(k, l) = g(k + l) + 9 h(k + e), (8.19) which is the same as Eq. (5.95).
Another way to derive this same result is to start with
C = 8, or Ch = Aß*,
Ck-1
(8.20)
in equation (8.9). Substituting this function in equation (8.9) and simplifying
gives
CqDe+1 – BDe + (p3²)De-1 = 0,
(8.21)
which has the characteristic equation
qr2 – Br + pB? = 0.
(8.22)
Therefore,
B,
PB
12 =
(8.23)
ri =
and
De = A3(B)B' + A4(B) ()
(8.24)
Since z(k, l, B) = Ch, De, we have
(k, l, 3) = Ã3(8)g*+te+ Ã4(8) ( 2)
(8.25)
where [A3 (3), A4 (B), A3(B), A4(B)] are arbitrary functions of B. If we now
sum/integrate over B, then
2(k, l) = g1(k +l) +
h1(k + l),
(8.26)
where gi and hị are functions of (k + l).
Let's now show that equations (8.19) and (8.26) are the same. To do this,
define a new discrete variable
т —
= k + l.
(8.27)
Therefore,
CO) - ()- ()" C)
k
m-l
m
(8.28)
and
k
т
h(k +l) =
h(k +l)
k+l
h(k + l)
h1(k + l).
(8.29)
This shows that equations (8.19) and (8.26) are equal.
Transcribed Image Text:Another way to derive this same result is to start with C = 8, or Ch = Aß*, Ck-1 (8.20) in equation (8.9). Substituting this function in equation (8.9) and simplifying gives CqDe+1 – BDe + (p3²)De-1 = 0, (8.21) which has the characteristic equation qr2 – Br + pB? = 0. (8.22) Therefore, B, PB 12 = (8.23) ri = and De = A3(B)B' + A4(B) () (8.24) Since z(k, l, B) = Ch, De, we have (k, l, 3) = Ã3(8)g*+te+ Ã4(8) ( 2) (8.25) where [A3 (3), A4 (B), A3(B), A4(B)] are arbitrary functions of B. If we now sum/integrate over B, then 2(k, l) = g1(k +l) + h1(k + l), (8.26) where gi and hị are functions of (k + l). Let's now show that equations (8.19) and (8.26) are the same. To do this, define a new discrete variable т — = k + l. (8.27) Therefore, CO) - ()- ()" C) k m-l m (8.28) and k т h(k +l) = h(k +l) k+l h(k + l) h1(k + l). (8.29) This shows that equations (8.19) and (8.26) are equal.
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