9.6.3 An Example To illustrate the use of the technique given in the previous section, consider the STDE +1+ y(2) = 0, (9.297) 4x2 (see equation (9.263)). For this case, 1 f(x) = 1+ 4x2 (9.298) and its discretization is fk = f(xk) = f(hk) 1 = 1+ (9.299) %3D 4h²k2 ° Using the Mickens-Ramadhani finite difference discretization, we obtain 1/27 2 cos (hVfk) 1 1+ 4h²k2 Yk+1 + Yk-1 Yk = 2 Cos |h Yk, (9.300) and this gives for ok the expression 1/27 V fk) 1 1 + 4h?k2 Ok = COS = cos Th (9.301)
9.6.3 An Example To illustrate the use of the technique given in the previous section, consider the STDE +1+ y(2) = 0, (9.297) 4x2 (see equation (9.263)). For this case, 1 f(x) = 1+ 4x2 (9.298) and its discretization is fk = f(xk) = f(hk) 1 = 1+ (9.299) %3D 4h²k2 ° Using the Mickens-Ramadhani finite difference discretization, we obtain 1/27 2 cos (hVfk) 1 1+ 4h²k2 Yk+1 + Yk-1 Yk = 2 Cos |h Yk, (9.300) and this gives for ok the expression 1/27 V fk) 1 1 + 4h?k2 Ok = COS = cos Th (9.301)
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 yellow
Transcribed Image Text:9.6.3
An Example
To illustrate the use of the technique given in the previous section, consider
the STDE
1
y" (x) + |1+
у (") — 0,
(9.297)
4x2
(see equation (9.263)). For this case,
1
f (x) = 1+
4.x2
(9.298)
and its discretization is
1
1+
4h?k2 °
fk = f("k) = f(hk):
(9.299)
Using the Mickens-Ramadhani finite difference discretization, we obtain
1/27
1
1+
4h²k²
Yk+1 + Yk-1= 2 |cos
Yk
= 2
h
CoS
Yk,
(9.300)
and this gives for
Ok
the expression
1
Ok = cos (h/fk)
= Cos |h.
(9.301)
4h2k2
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