1690 AC FAB-0.7809FBC 0 FCD+0.8575F, = 0 DE 0.9231F 0.3846F AC 0.6247FBC-FBD = 0 FBD-0.5145F Fr -FAB DE 0.3846FCE-0.3846F AC-0.7809FBC-FCD = 0 0.9231FAC+0.6247FBC-0.9231FCE = 0 = 3625 DF 0

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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solve using gauss seidel

0.9231F
1690
FAB-0.7809 FBC 0
=
AC
=
FCD+0.8575F, 0
DE
-FAB
0.3846F AC
0.6247FBC FBD = 0
- FDF
FBD-0.5145F DE
0.3846 FCE-0.3846F AC-0.7809FBC-FCD = 0
0.9231 FAC+0.6247 FBC-0.9231FCE = 0
3625
= 0
Transcribed Image Text:0.9231F 1690 FAB-0.7809 FBC 0 = AC = FCD+0.8575F, 0 DE -FAB 0.3846F AC 0.6247FBC FBD = 0 - FDF FBD-0.5145F DE 0.3846 FCE-0.3846F AC-0.7809FBC-FCD = 0 0.9231 FAC+0.6247 FBC-0.9231FCE = 0 3625 = 0
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