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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.
Download (direct link): elementarydifferentialequations2001.pdf
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9 2.684604 5.419209 2.968276 5.936551 2.968284
10 2.968284
From Eq.(10) we have h
+1 - + ------ (9fn+1 + 19fn - 5fn-1 + fn-2)
24
- + ------ [9(.5
24
-+1
+ 2+1) + 19fn - 5f
-1
+ f
n-2
].
Solving for yn+i we obtain h
+1 - [ + --- (19fn - 5fn-1 + fn-2 + 4.5 - 9tn+1)]/(1-.75h).
24
For h - .05, t0 - 0, y0 - 1 and using y1 and y2 as calculated using the Runge-Kutta formula, we obtain the following table:
n fn +1
0 1 2.5
1 1.130171 2.710342
2 1.271403 2.942805 1.424859
1.424859 3.199718 1.591825
4 1.591825 3.483650 1.773722
5 1.773722 3.797444 1.972120
6 1.972120 4.144241 2.188755
7 2.188755 4.527510 2.425544
8 2.425544 4.951088 2.684607
9 2.684607 5.419214 2.968287
10 2.968287
From Eq (16) we have
+1 - (48 - 36-1 + 16-2 - 3-3 + 12hfn+1)/25
- [48yn - 36-1 + 16yn-2 - - + 12h(.5-tn+1)]/25+(24/25)hyn+1. Solving for yn+1 we have
+1 - [48yn - 36-1 + 16yn-2 - - + 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) = -^ P2 (tn) = ^ P2 (tn+1) = +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 + h) + C = -1 (i)
-hn
t
B(
2n + B + C = yn (ii)
t ft
A n
A(tn+h) 2 + h) + C = +1 (iii:
+hn
t
B(
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+1 - 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 +1 = (1/3)(4yn-yn-1+2hfn+1), which is Eq. (15).
Section 8.5, Page 4 54
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 f1(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/(c1e-t-1)]. Setting t = 0 and y = 0, we obtain c1 = 2 and thus f 1(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
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