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

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5.0

2.24

and for h = .18 we obtain:

1

t = 4.14 4.32 4.50 4.68 4.86 5.04

y = 4.26 4.68 5.04 5.48 5.89 6.35.

Clearly the second set of values is stable, although far from accurate.

Section 8.5

171

4c. For a step size of .25 we find

t - 4 4.25 4.75 5.00

y - 4.018 4.533 5.656 6.205

for a step size of .28 we find

t - 4.2 4.48 4.76 5.00

y - 10.14 10.89 11.68 12.51

and for a step size of .3 we find

t - 4.2 y - 353

4.5

484

4.8

664

5.1

912.

Thus instability appears to occur about h - .28 and certainly by h - .3. Note that the exact solution for t - 5 is y - 6.2500, so for h - .25 we do obtain a good approximation.

4d. For h - .5 the error at t - 5 is .013, while for h - .385,

the error at t - 5.005 is .01.

5a. The general solution of the D.E. is y(t) - t + ce1t, where y(0) - 0 ^ c - 0 and thus y(t) - t, which is independent of l.

5c. Your result in Part b will depend upon the particular

computer system and software that you use. If there is sufficient accuracy, you will obtain the solution y - t for t on 0 < t < 1 for each value of l that is given,

since there is no discretization error. If there is not

sufficient accuracy, then round-off error will affect your calculations. For the larger values of l, the numerical solution will quickly diverge from the exact solution, y - t, to the general solution y - t + ce1t, where the value of c depends upon the round-off error.

If the latter case does not occur, you may simulate it by computing the numerical solution to the I.V.P. y' - ly - 1 - It, y(.1) - .10000001. Here we have assumed that the numerical solution is exact up to the point t - .09 [i.e. y(.09) - .09] and that at t - .1 round-off error has occurred as indicated by the slight error in the I.C. It has also been found that a larger step size (h - .05 or h - .1) may also lead to round-off error.

172

Section 8.6

Section 8.6, Page 457

2a. The Euler formula is xn+1 = xn + hfn,

where

^2xn+tnYnA

fi =

xnYn

¥

, x0 = 1 and y0 = 1.Thus f0 =

2-0 (1)(1) ó

1+.1(2) 1+.1(1)

1.2

1.1

+ 1 ( 2. 51 ^

4 1

CN 1

v (1.2)(1.1) y v 1. 32 ó

f1.2+.1(2.51) 1 f1.4511 ff(.2) N

X2 = =

¨1.1+.1(1.32) v1.232 y vV(.2) ó

2b. Eqs. (7) give:

k01 =

1 ' 2 + 0 1 V 2 1

1

1 , (1)(1) 7 v 1y

1

k

k

02

03

'2.51^

V1.32 ó \

2.4+.1(1.1)

(1.2)(1.1)

2.502+.1(1.132)

(1.251)(1.132)

3.04608+.2(1.28323)

(1.52304)(1.28323)

Using Eq. (6) in scalar form, we then have

x1 = 1+(.2/6)[2+2(2.51)+2(2.6152)+3.30273] = 1.51844

Ó! = 1+(.2/6)[1+2(1.32)+2(1.41613)+1.95440] = 1.28089.

k04 =

2.6152

1.41613

3.30273 1.95441

fn =

x

1

1 I

7. Write a computer program to do this problem as there are twenty steps or more for h < .05.

8. If we let y = x', then y' = x" and thus we obtain the system x' = y and y' = t-3x-t2y, with x(0) = 1 and y(0) = x'(0) = 2. Thus f(t,x,y) = y,

g(t,x,y) = t - 3x - t2y, t0 = 0, x0 = 1 and y0 = 2. If a program has been written for an earlier problem, then its best to use that. Otherwise, the first two steps are as follows:

k01 =

2

-3

Section 8.6

173

ê02 -

"03

04

2+(-.15)

.05-3(1.1) — (.05) 2(1.85)

2+(—.16273)

.05—3(1.0925)—(.05)2(1.83727) 2+(—.32321)

.1—3(1.18373) —(.1) 2(1.67679)

1.85

v—3.25463 ó

Ë /

\ ã

1.83727

—3.23209

1.67679 ó V—3.46796 ó

and thus

õ1 - 1+(.1/6)[2 + 2(1.85)+2(1.83727)+(1.67679)]-1.18419, ó 1 - 2+(.1/6)[-3-2(3.25463)-2(3.23209)-3.46796]-1.67598,

which are approximations to x(.1) and y(.1) - x'(.1).

In a similar fashion we find

êö -

1.67598

—3.46933

ê12 -

1.50251

—3.68777

k13 -

1.49159

V—3.66151 ó and thus

k14 -

—3.85244

x2-x1+(.1/6)[1.67598+2(1.50251)+2(1.49159)+1.30983]-1.33376 y2-y1—(.1/6)[3.46933+2(3.68777)+2(3.66151)+3.85244]-1.30897.

Three more steps must be taken in order to approximate x(.5) and y(.5) - x'(.5). The intermediate steps yield x(.3) @ 1.44489, y(.3) @ .9093062 and x(.4) @ 1.51499, y(.4) @ .4908795.

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