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# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
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0
k
4 - f( .1, - 2 + . . 1k03) - f(.1, - 1.915202) - . 897927, and
0
k
Ó1 - -2 + .1(k01 + 2k02 + 2k03 + k04)/6 - -1. 915221. For
comparison, see Problem 11 in Sections 8.1 and 8.2.
166
Section 8.4
14a.
14b. We have f(tn,yn) = tn
+ óÏ.
t0
o, y0
1 and h
.1 so
k01
k02
o2 + 12
= (.05)
k.
03
1
+ (1 + .05)2 = 1.105
= (.05)2 + [1 + .05(1.105)]2 = 1.11605
k04 = (.1)2 + [1 + .1(1.11605)]2 = 1.245666 and thus
Ó! = 1 + .1(k01 + 2k02 + 2k03 + k04)/6 = 1.11146. Using these steps in a computer program, we obtain the following values for y:
h = .025 5.8483 14.3021 49.757
14c. No accurate solution can be obtained for y(1), as the values at t = .975 for h = .025 and h = .0125 are 1218 and 23,279 respectively. These are caused by the slope field becoming vertical as t ^ 1.
t
.8
.9
.95
h =.1 5.842 14.0218
h = .05 5.8481 14.2712 46.578
h = 0.125 5.8486 14.3046 50.3935
Section 8.4, Page 444
4a. The predictor formula is
Óï+1 = Yn + h(55fn - 59fn-1 + 37fn-2 - 9f-3)/24 and the corrector formula is
Óï+1 = Yn + h(9fn+1 +19fn -5fn-1 + fn-2)/24, where
fn = 2tn + exp(-tnyn). Using the Runge-Kutta method, from Section 8.3, Problem 4a, we have for t0 = 0 and y0 = 1,
Ó! = 1.1048431, y2 = 1.2188411 and y3 = 1.3414680. Thus the predictor formula gives y4 = 1.4725974, so f4 = 1.3548603 and
the corrector formula then gives y4 = 1.4726173, which is
desired value. These results, and the next step, are
summarized in the following e:
l
b
a
t
n Óï fn Óï+1 fn+1 Óï+1
0 1 1 Corrected
1 1.1048431 1.0954004
2 1.2188411 1.1836692
3 1.3414680 1.2686862 1.4725974 1.3548603 1.4726173
4 1.4726173 1.3548559 1.6126246 1.4465016 1.6126215
5 1.6126215
Section 8.4
167
where fn is given above, yn+1 is given by the predictor formula, and the corrected yn+1 is given by the corrector formula. Note that the value for f4 on the line for n = 4 uses the corrected value for y4, and differs slightly from the f4 on the line for n = 3, which uses the predicted value for y4.
4b. The fourth order Adams-Moulton method is given by
Eq. (10): Óï+1 = Óï + (h/24)(9f+1 + 19fn - 5f-1 + fn-2). Substituting h = .1 we obtain
Óï+1 = Óï + (.0375)(19fn - 5fn-1 + fn-2) + .0375fn+1.
For n = 2 we then have
Ó3 = Ó2 + (.0375)(19f2 - 5f1 + f0) + .0375f3
= 1.293894103 + .0375(.6 + e".3y3), using values for
y2, f0, f1, f2 from part a. An equation solver then yields y3 = 1.341469821. Likewise
y4 = y3 + (.0375)(19f3 - 5f2 + f1) + .0375f4
= 1.421811841 + .0375(.8 + e-.4y4), where f3 is calculated
using the y3 found above. This last equation yields y4 = 1.472618922. Finally
y5 = y4 + (.0375)(19f4 - 5f3 + f2) + .0375f5
= 1.558379316 + .0375(1.0 + e".5y5), which gives y5 = 1.612623138.
4c. We use Eq. (16):
Óï+1 = (1/25)(48Óï - 36Óï-1 + 16Ón-2 - 3Óï-3 + 12hfn+1).
Thus y4 = .04(48y3 - 36y2 + 16y1 - 3y0) + .048f4
= 1.40758686 + .048(.8 + e".4y4), using values
for y0, y1. y2, y3 from part a. An equation solver then yields y4 = 1.472619913. Likewise y5 = .04(48y4 - 36y3 + 16y2 - 3y1) + .048f5
= 1.54319349 + .048(1 + e .5y5), which gives y5 = 1.612625556.
7a. Using the predictor and corrector formulas (Eqs.6 and 10) with fn = .5 - tn + 2yn and using the Runge-Kutta method to calculate y1,y2 and y3, we obtain the following table for h = .05, t0 = 0, y0 = 1:
168
Section 8.4
7b.
7c.
n Óï fn Óï+1 fn+1 Óï+1
corrected
0 1 2.5
1 1.130171 2.710342
2 1.271403 2.9420805
3 1.424858 3.199717 1.591820 3.483640 1.591825
4 1.591825 3.483649 1.773716 3.797433 1.773721
5 1.773721 3.797443 1.972114 4.144227 1.972119
6 1.972119 4.144238 2.188747 4.527495 2.188753
7 2.188753 4.527507 2.425535 4.951070 2.425542
8 2.425542 4.951084 2.684597 5.419194 2.684604
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