5. Solve the following initial value problem (IVP) y(0) = | '– xy = x by using (a) second-order Taylor's series method with h = 0.2 and 0sxs1. (b) second-order Taylor's series method with h = 0.1, 0.25, 0.5 and 0 s xs1 (c) third-order Taylor's series method with h = 0.2,0.25,0.5 and 0 sx<1. Hence, if the exact solution is y = 2eT -1, find its errors.
5. Solve the following initial value problem (IVP) y(0) = | '– xy = x by using (a) second-order Taylor's series method with h = 0.2 and 0sxs1. (b) second-order Taylor's series method with h = 0.1, 0.25, 0.5 and 0 s xs1 (c) third-order Taylor's series method with h = 0.2,0.25,0.5 and 0 sx<1. Hence, if the exact solution is y = 2eT -1, find its errors.
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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Question
By using formula second order Taylor series method solve the question 5 .
Transcribed Image Text:5. Solve the following initial value problem (IVP)
'–xy = x y(0) =
by using
(a) second-order Taylor's series method with h = 0.2 and 0<xs1.
(b) second-order Taylor's series method with h = 0.1,0.25,0.5 and 0<x<1
(c) third-order Taylor's series method with h = 0.2,0.25,0.5 and 0<x<1.
Hence, if the exact solution is y = 2eT -1, find its errors.
ANSWER:
5. (a)
h= 0.5
error|
errot
exact
exact
1.000
1.000
1.000
1.000
1
0.2
1.040
1.040
0.000
0.5
1.0
1.250
2.164
1.266
2.297
0.016
0.133
1.167
1.394
0.003
0.007
0.4
1.164
1.387
0.6
4
0.8
1.0
1.738
2.266
1.754
0.016
0.031
(c) h= 0.2
2.297
Jerror|
exact
1.000
1.000
(b) h =0.1
0.2
1.040
1.166
1.393
1.040
0.000
error|
exact
0.4
3
0.6
1.167
0.001
0.001
0.002
0.004
1.000
1.000
1.394
1.010
1.040
1.091
1
0.1
1.010
0.000
4
0.8
1.752
1.754
0.2
3
0.3
1.040
1.092
1.167
0.000
0.001
1.0
2.293
2.297
4
0.4
1.165
0.002
h = 0.25
0.5
0.6
1.264
1.391
1.266
1.394
0.002
0.003
Jerrot|
exact
1.000
1.000
7
0.7
0.8
1.551
1.749
1.991
1.555
1.754
1.999
0.004
0.005
0.25
0.50
0.75
1.062
1.263
1.643
1.063
1.266
1.650
0.001
0.003
0.007
0.9
0.008
10 1.0
2.287
2.297
0.010
3
4
1.00
2.285
2.297
0.012
h-0.25
h= 0.5
errot|
exact
Jerrot|
X,
exact
1.000
1.000
1
0.25
1.062
1.063
0.001
1.000
1.000
1
0.5
1.0
1.250
2.241
1.266
2.297
0.50
1.259
1.266
0.007
0.016
3
0.75
1.629
1.650
0.021
0.056
4
1.00
2.249
2.297
0.048
12
123
in
no790 a
Transcribed Image Text:Formula
Second Order Taylor Series Method
y(x) = y(x;) + hy'(x;) + "(x,)
When the Taylor's series is truncated after three terms, it is called second order
Taylor's series method. Else, we write as
Vis = y; + hy +
2!
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