where C(u) is an arbitrary function of µ. Now, k 2µ (2µ + 3)* = 3k ( 1+ 3 2 = 3k + k- 3 2µk(k- 1) (2µ (5.106) 2! 3 ()" - (4)]: k-1 2µ +...+k 3 3 and, therefore, equation (5.105) can be written z(k, e) = 3*[H(e) + 2/3kH(l + 1) + 2/ok(k – 1)H(l+2) +...+ (2/3)* H(l + k)], (5.107) where H(l) is an arbitrary function of l and we have used the results

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.10: Partial Fractions
Problem 12E
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5.3.5 Еxample E
Applying the method of Lagrange to the partial difference equation
z(k + 1, €) – 2z(k, l+ 1) – 3z(k, l) = 0
(5.103)
LINEAR PARTIAL DIFFERENCE EQUATIONS
179
gives A = 2µ
3, and special solutions having the form
zp(k, l) = (2µ +3)*µº.
(5.104)
General solutions can be obtained by summing over u; doing this gives
z(k, l) = C(H)(2µ +3)*µ",
(5.105)
where C(u) is an arbitrary function of µ. Now,
k
2µ
(2µ + 3)* = 3* (1+
3
k(k – 1)
3k
(5.106)
3
2!
3
(#)]
k-1
2µ
+...+ k
+
3
3
and, therefore, equation (5.105) can be written
z(k, l) = 3*[H(e) + 2/3kH(l + 1) + 2/9k(k – 1)H(l+2)
+...+ (2/3)* H(l + k)],
(5.107)
where H(l) is an arbitrary function of l and we have used the results
H(€) = £C(w)µ°,
(5.108)
H(l + m) = C(H)µ²+m.
Applying the method of separation of variables gives, for Ck and De, the
equations
Ck+1
2De+1+ 3De
= a,
(5.109)
De
the solutions of which are
Ck = Aa", De = B
(5.110)
where A and B are arbitrary constants. If we let
a – 3
(5.111)
then z(k, l) = CrDe becomes
z(k, l) = AB(3+ 2µ)*µ°,
(5.112)
which is the same as equation (5.104). Therefore, the separation-of-variables
method gives the same solution as presented in equation (5.107).
Transcribed Image Text:5.3.5 Еxample E Applying the method of Lagrange to the partial difference equation z(k + 1, €) – 2z(k, l+ 1) – 3z(k, l) = 0 (5.103) LINEAR PARTIAL DIFFERENCE EQUATIONS 179 gives A = 2µ 3, and special solutions having the form zp(k, l) = (2µ +3)*µº. (5.104) General solutions can be obtained by summing over u; doing this gives z(k, l) = C(H)(2µ +3)*µ", (5.105) where C(u) is an arbitrary function of µ. Now, k 2µ (2µ + 3)* = 3* (1+ 3 k(k – 1) 3k (5.106) 3 2! 3 (#)] k-1 2µ +...+ k + 3 3 and, therefore, equation (5.105) can be written z(k, l) = 3*[H(e) + 2/3kH(l + 1) + 2/9k(k – 1)H(l+2) +...+ (2/3)* H(l + k)], (5.107) where H(l) is an arbitrary function of l and we have used the results H(€) = £C(w)µ°, (5.108) H(l + m) = C(H)µ²+m. Applying the method of separation of variables gives, for Ck and De, the equations Ck+1 2De+1+ 3De = a, (5.109) De the solutions of which are Ck = Aa", De = B (5.110) where A and B are arbitrary constants. If we let a – 3 (5.111) then z(k, l) = CrDe becomes z(k, l) = AB(3+ 2µ)*µ°, (5.112) which is the same as equation (5.104). Therefore, the separation-of-variables method gives the same solution as presented in equation (5.107).
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