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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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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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