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

**Download**(direct link)

**:**

**475**> 476 477 478 479 480 481 .. 609 >> Next

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

**475**> 476 477 478 479 480 481 .. 609 >> Next