5.5.3 Example C The equation z(k +1, l + 1) + 2z(k, l) = 3*(k² + l + 1) (5.176) corresponds to equation (5.158) with the following identifications: 6(E1, E2) = EL E2+2, A = 3, a = 1, b = 0, and F(k, e) = k2 +l+1. From equation (5.159) we see that the solution z(k, l) can be expressed as z(k, l) = 3*w(k, l), (5.177) where w(k, l) satisfies the equation (3E1E2 + 2)w(k, l) = k² + l + 1. (5.178) The particular solution to this equation can be obtained as follows: Wp(k, l) 1 -(k² +l+1) 2+ 3E1E2 1 1 - (k² + l+1) 51+ 3/5(A1 + A2 + A¡A2) = 1s[1 – 3/(A1 + A2 + A¡A2) +%/25(△? + △2 + 스스 + .)+…J(k2 + l+ 1) = 1/125(25k2 – 30k + 25l + 13). (5.179) %3D %3D Likewise, the homogeneous equation
5.5.3 Example C The equation z(k +1, l + 1) + 2z(k, l) = 3*(k² + l + 1) (5.176) corresponds to equation (5.158) with the following identifications: 6(E1, E2) = EL E2+2, A = 3, a = 1, b = 0, and F(k, e) = k2 +l+1. From equation (5.159) we see that the solution z(k, l) can be expressed as z(k, l) = 3*w(k, l), (5.177) where w(k, l) satisfies the equation (3E1E2 + 2)w(k, l) = k² + l + 1. (5.178) The particular solution to this equation can be obtained as follows: Wp(k, l) 1 -(k² +l+1) 2+ 3E1E2 1 1 - (k² + l+1) 51+ 3/5(A1 + A2 + A¡A2) = 1s[1 – 3/(A1 + A2 + A¡A2) +%/25(△? + △2 + 스스 + .)+…J(k2 + l+ 1) = 1/125(25k2 – 30k + 25l + 13). (5.179) %3D %3D Likewise, the homogeneous equation
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 blue
Transcribed Image Text:5.5.3 Example C
The equation
z(k +1, l +1) + 2z(k, l) = 3* (k² + l + 1)
(5.176)
corresponds to equation (5.158) with the following identifications: ø(E1, E2) =
E, E2 +2, A = 3, a = 1, b = 0, and F(k, l) = k² +l+1. From equation (5.159)
we see that the solution z(k, l) can be expressed as
z(k, l) = 3* w(k, l),
(5.177)
where w(k, l) satisfies the equation
(3E1E2 + 2)w(k, l) = k² + l + 1.
(5.178)
The particular solution to this equation can be obtained as follows:
>(k,e)
1
(k² + l + 1)
2+ 3E1E2
1
1
-(k² + l+1)
51+ 3/5(A1 + A2 + A¡A2)
= 15[1 – 3/5(A1 + A2 + A¡A2)
+9/25(△? + △2 + 스스 + )+.(k2 + l+ 1)
= 1/125(25k² – 3Ok + 25l + 13).
(5.179)
Likewise, the homogeneous equation
(3E1E2 + 2)wn (k, l) = 0
(5.180)
has the solution
Wh (k, l) = (-2/3)“ f(l – k),
(5.181)
where f is an arbitrary function of l – k. Therefore, the complete solution of
equation (5.178) is given by the sum of the expressions in equations (5.179)
and (5.181). Finally, putting these results into equation (5.177) gives
z(k, e) = (-2)* f(l – k) +
3k
(25k2 – 30k + 25l + 13).
125
(5.182)
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