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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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= [48Yn - 36Yn-1 + 16Yn-2 - 3Yn-3 + 12h(.5-tn+1)]/25+(24/25)hYn+1. Solving for Yn+1 we have
Yn+1 = [48Yn - 36Yn-1 + 16Yn-2 - 3Yn-3 + 12h(.5-tn+1)]/(25-24h). Again, using Runge-Kutta to find y1 and y2, we then obtain the following table:
Section 8.5
169
n yn yn+1
0 1
1 1. 130170833
2 1. 271402571
3 1. 424858497 1. 591825573
4 1. 591825573 1. 773724801
5 1. 773724801 1. 972125968
6 1. 972125968 2. 188764173
7 2. 188764173 2. 425557376
8 2. 425557376 2. 684625416
9 2. 684625416 2. 968311063
10 2. 968311063
The exact solution is y(t) = et + t/2 so y(.5) = 2.9682818 and y(2) = 55.59815, so we see that the predictor-corrector method in part a is accurate through three decimal places.
16. Let P2(t) = At2 + Bt + C. As in Eqs. (12) and (13) let
P2 (tn-1) = yn-^ P2 (tn) = y^ P2 (tn+1) = yn+1 and
P2(tn+1) = f(tn+1,yn+1) = fn+1. Recall that tn-1 = tn -h
and tn+i = tn + h and thus we have the four equations:
A(tn-h) 2 + B(tn-h) + C = yn-1 (i)
At2 + Btn + C = yn (ii)
A(tn+h) 2 + B(tn+h) + C = yn+1 (iii!
2A(tn+h) + B = fn+1 (iv)
Subtracting Eq. (i) from Eq. (ii) to get Eq. (v) (not shown) and subtracting Eq. (ii) from Eq. (iii) to get Eq. (vi) (not shown), then subtracting Eq. (v) from Eq. (vi) yields yn+i - 2yn + yn-1 = 2Ah2, which can be solved for A. Thus B = fn+1 - 2A(tn+h) [from Eq. (iv)] and C = yn - tnfn+1 + Atn + 2Atnh [from Eq. (ii)]. Using these values for A, B and C in Eq. (iv) yields yn+i = (1/3)(4yn-yn-i+2hfn+i), which is Eq. (15).
Section 8.5, Page 454
2a. If 0 ? t ? 1 then we know 0 ? t2 ? 1 and hence
ey ? t2 + ey ? 1 + ey. Since each of these terms represents a slope, we may conclude that the solution of Eq.(i) is bounded above by the solution of Eq.(iii) and is bounded below by the solution of Eq.(iv).
2b. f1(t) and f2(t) can each be found by separation of
170
Section 8.5
variables. For fi(t) we have
1 + ey
dy = dt, or
>-y
dy = dt. Integrating both sides yields
e-y+1
-ln(e-y+1) = t + c. Solving for y we find
y = ln[1/(cxe-t-1)]. Setting t = 0 and y = 0, we obtain cx = 2 and thus f i(t) = ln[et/(2-et)]. As t ^ ln 2, we see that f1(t) ^ . A similar analysis shows that f2(t) = ln[1/(c2-t)], where c2 = 1 when the I.C. are used. Thus f 2(t) ^ as t ^ 1 and thus we conclude that f(t) ^ for some t such that ln2 ? t ? 1.
2c. From Part b: f1(.9) = ln[1/(^e -1)] = 3.4298 yields
c1 = 2.5393 and thus f1(t) ^ when t @ .9319. Similarly for f2(t) we have c2 = .9324 and thus f2(t) ^ when t @ .932.
-10t
is the
undetermined coefficients, is
4a. The D.E. is y' + 10y = 2.5t2 + .5t. So yh = ce
solution of the related homogeneous equation and the particular solution, using yp = At2 + Bt + C.
Substituting this into the D.E. yields A = 1/4, B = C = 0.
To satisfy the I.C., c = 4, so
y(t) = 4e-10t + (t2/4), which is shown in the graph.
4b. From the discussion following Eq (15), we see that h must
2
be less than
Irl
for the Euler method to be stable.
Thus, for r = 10, h < following values:
For h = .2 we obtain the
t = 4
y = 8
4.2
.4
4.4
8.84
4.6
1.28
4.8
9.76
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 = 6.25oo, so for h = .25 we do obtain a good approximation.
4d. For h = .5 the error at t = 5 is .o13, while for h = .385,
the error at t = 5.oo5 is .o1.
5a. The general solution of the D.E. is y(t) = t + ce1t, where y(o) = o ^ c = o 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 o ? 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) = .1oooooo1. Here we have assumed that the numerical solution is exact up to the point t = .o9 [i.e. y(.o9) = .o9] 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 = .o5 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,
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