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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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n yn tn +1
0 1 0 1 .06513
1 1 .06513 025 1 .13303
2 1 .13303 050 1 .20381
38 7 .49768 950 7 .87980
39 7 .87980 975 8 .28137
40 8 .28137 1. 000 8 .70341
78 55. 62105 1 .950 58. 50966
79 58. 50966 1 .975 61. 54964
80 61. 54964 2 .000
least eight decimal pl aces were us ed in
calculations.
9c. The backward Euler formula gives
yn+1 = yn + hy/ tn+1 + yn+1 . Subtracting yn from both sides, squaring both sides, and solving for yn+1 yields
yn+1 = yn +
+ hVyn + tn+1 + h2/4 . Alternately, an
2
15.
equation solver can be used to solve
yn+1 = yn + hyj tn+1 + yn+1 for yn+1. The first few values, for h = 0.25, are y1 = 3.043795, y2 = 2.088082, y3 = 3.132858 and y4 = 3.178122 @ y(.1).
If y' = 1 - t + 4y then
y" = -1 + 4y' = -1 + 4(1-t+4y) = 3 - 4t + 16y. In
Eq.(12) we let yn, y'n and denote the approximate
values of f(tn), f'(tn), and f"(tn), respectively.
Keeping the first three terms in the Taylor series we have
+1 = + y'nh + h2/2
= + (1 - tn + 4yn)h + (3 - 4tn + 16yn)h2/2. For n = 0,
t0 = 0 and y0 = 1 we have
y 1 = 1 + (1 - 0 + 4)(.1) + (3 - 0 + 16)
(.1)
2
= 1.595.
16. If y = f(t) is the exact solution of the I.V.P., then f'(t) = 2f(t) - 1 and f"(t) = 2f'(t) = 4f(t) - 2. From
Eq.(21), en+1 = [2f(tn) - 1]h2, tn < tn < tn + h. Thus a
2
162
Section 8.1
bound for en+1 is |en+1| < [1 + 2 max |f (t)|]h . Since the
exact solution is y = f(t) = [1 + exp(2t)]/2,
en+1 = h2exp(2tn). Therefore |e1| < (0.1)2exp(0.2) = 0.012 and
|e4| < (0.1)2exp(0.8) = 0.022, since the maximum value of
exp(2tn) occurs at t = .1 and t = .4 respectively. From Problem 2 of Section 2.7, the actual error in the first step is .0107.
19. The local truncation error is en+1 = f"(tn)h2/2. For this problem f'(t) = 5t - 3f 1/2(t) and thus f"(t) = 5 - (3/2) f _1/2f' = 19/2 - (15/2)tf-1/2.
Substituting this last expression into en+1 yields the desired answer.
22d. Since y" = -5psin5pt, Eq.(21) gives en+1 = -(5p/2)sin(5ptn)h2
I I 5p 2 1
Thus I en+1 I < ----h < .05, or h < @ .08.
2 V50p
23a. From Eq.(14) we have En = f(tn) - yn. Using this in
Eq.(20) we obtain
En+1 = En + h{f[tn,f(tn)] - f(tn,yn)} + f"(tn)h2/2. Using
the given inequality involving L we have
|f[tn,f(tn)] - f(tn,yn)| < L |f(tn) - yn| = L|En| and thus
|En+1| < |En| + hL|En| + max |f"(t)|h2/2 = a|En| + Ph2.
23b. Since a = 1 + hL, a - 1 = hL. Hence bh2(an-1)/(a-1) = bh2 [(1+hL) n - 1]/hL = bh[(1+hL) n - 1]/L.
23c. (1+hL)n < exp(nhL) follows from the observation that exp(nhL) = [exp(nL)]n = (1 + hL + h2L2/2! + ...)n.
Noting that nh = tn - t0, the rest follows from Eq.(ii).
24. The Taylor series for f(t) about t = tn+1 is
(t - tn+ 1 ) 2
f(t) = f (tn+1) + f7 (tn+1)(t-tn+1) + f"(tn+1) + .
2
Letting f'(t) = f(t,f(t)), t = tn and h = tn+1 - tn we have f (tn) = f (tn+1) - f(tn+1,f (tn+1))h + f"(tn)h2/2, where tn < tn < tn+1. Thus
f (tn+1) = f (tn) + f(tn+1,f (tn+1))h - f"(tn)h2/2. Comparing
this to Eq. 13 we then have en+1 = -f"(tn)h2/2.
Section 8.2
163
2 5b. From Problem 1 we have yn+1 = yn + h(3 + tn - yn), so y 1 = 1 + .05(3 + 0 - 1) = 1.1
y2 = 1.1 + .05(3 + .05 - 1.1) = 1.20 @ y(.1)
y3 = 1.20 + .05(3 + .1 - 1.20) = 1.30
y4 = 1.30 + .05(3 + .15 - 1.30) = 1.39 @ y(.2).
Section 8.2, Page 434
1a. The improved Euler formula is
+1 = yn + [yn + f(tn + h, yn + hyn)]h/2 where
y' = f(t,y) = 3 + t - y. Hence = 3 + tn - yn and
+ h, + hyP) = 3 + tn+1 - (y
n
y
= yn + (3 + tn - yn)h + h2
--- (1 -
2
= yn + (3 + tn - yn)h + h2
- (-2
2
1 + (3-1) (.05) (.05) 2 (-2+1)
+
Thus
2
(.05)2
2 = ӳ + (3 +.05 - yi)(.05) + ---------- (-2 -.05 + yi) = 1.19512
2
are the first two steps. In this case, the equation
specifying +1 is somewhat more complicated when
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