Equation (5.119) rewritten with the assumed form z(k, l) = CkDe is Ck+3 De – 3Ck+2De+1+3Ck+1De+2 – CkDe+3 = 0. (8.30) Dividing all terms by Ck Dk gives ((똥)- (똥) () Ck+3 Ck+2 Ck+1 +3 De+2 De De+1 De+3 De = 0. (8.31) We can start with either one of the following choices Ck+1 De+1 or (8.32) = a. Ck De Using the second expression gives De = Aa', A = arbitrary function of a. (8.33) %3D Substitution of this result in equation (8.31) yields Ck+3 Ck+2 3a + 3a? Ck (Ot1) - a' = 0, Ck+1 (8.34) Co or Ck+3 – (3a)Ck+2 + 3a²Ck+1 – a³Ch = 0. (8.35) The corresponding characteristic equation is p3 – (3a)r2 + (3a²)r – a³ = (r – a)³ = 0. (8.36) | Therefore, C(a) is Ck(a) = A1(a)a* + A2(a)ka* + A3(@)k²a* (8.37) and z(k, l, a) = Cr(@)D¿(a) %3D = á(a)a*+l + Ã2(a)ka*+l + Ã3(a)k²a*+e. (8.38) If we sum/integrate over a, we obtain the following solution to equation (5.119) 2(k, e) = f(k + l) + kg(k + l) + k²h(k +l), (8.39) where (f, g, h) are arbitrary functions of (k + e).
Equation (5.119) rewritten with the assumed form z(k, l) = CkDe is Ck+3 De – 3Ck+2De+1+3Ck+1De+2 – CkDe+3 = 0. (8.30) Dividing all terms by Ck Dk gives ((똥)- (똥) () Ck+3 Ck+2 Ck+1 +3 De+2 De De+1 De+3 De = 0. (8.31) We can start with either one of the following choices Ck+1 De+1 or (8.32) = a. Ck De Using the second expression gives De = Aa', A = arbitrary function of a. (8.33) %3D Substitution of this result in equation (8.31) yields Ck+3 Ck+2 3a + 3a? Ck (Ot1) - a' = 0, Ck+1 (8.34) Co or Ck+3 – (3a)Ck+2 + 3a²Ck+1 – a³Ch = 0. (8.35) The corresponding characteristic equation is p3 – (3a)r2 + (3a²)r – a³ = (r – a)³ = 0. (8.36) | Therefore, C(a) is Ck(a) = A1(a)a* + A2(a)ka* + A3(@)k²a* (8.37) and z(k, l, a) = Cr(@)D¿(a) %3D = á(a)a*+l + Ã2(a)ka*+l + Ã3(a)k²a*+e. (8.38) If we sum/integrate over a, we obtain the following solution to equation (5.119) 2(k, e) = f(k + l) + kg(k + l) + k²h(k +l), (8.39) where (f, g, h) are arbitrary functions of (k + e).
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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Question
Explain the determine yellow and the equation 5.119 is here
Transcribed Image Text:The equation
2(k + 3, l) – 3z(k + 2, l+ 1) + 3z(k +1, l+ 2) – 2(k, l + 3) = 0
(5.119)
cannot be solved by the method of separation of variables. However, La-
Transcribed Image Text:8.2.2
Example B
Equation (5.119) rewritten with the assumed form z(k, l) = Ch De is
Ck+3 De – 3Ck+2De+1+ 3Ck+1De+2 - Ck De43
= 0.
(8.30)
Dividing all terms by Cr Dk gives
´Ck+3
´Ck+2
-3
De+1
´Ck+1
+3
De+2
De+3
= 0. (8.31)
Ck
De
Ck
De
De
We can start with either one of the following choices
Ck+1
De+1
= a
Ck
(8.32)
or
= a.
De
Using the second expression gives
De = Aa", A = arbitrary function of a.
(8.33)
Substitution of this result in equation (8.31) yields
a () ()
´Ck+1
Ck+3
Ck
Ck+2
За
- a³ = 0,
+ 3a?
(8.34)
or
Ck+3 – (3a)Ck+2 + 3a²Ck+1 – a³Ck = 0.
(8.35)
The corresponding characteristic equation is
p3 – (3a)r2 + (3a²)r – a³ = (r – a)³ = 0.
(8.36)
Therefore, C (a) is
Ch(a) = A1(a)a* + A2(@)ka* + A3(a)k²a*
(8.37)
and
z(k, l, a) = Cr(@)De(a)
= á(a)a*+l + Ã2(a)ka*+e + Ã3(a)k²a*+e.
(8.38)
If we sum/integrate over a, we obtain the following solution to equation
(5.119)
C2(k, l) = f(k + l) + kg(k + l) + k²h(k+ l),
(8.39)
where (f, g, h) are arbitrary functions of (k + e).
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