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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
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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+i = yn + h(3 + tn - yn), so yi = 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
yn+1 = yn + [yn + f(tn + h, yn + hyn)]h/2 where
y' = f(t,y) = 3 + t - y. Hence yn = 3 + tn - yn and
+ h, yn + hyn) = 3 + tn+1 - (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
y2 = y1 + (3 +.05 - y1)(.05) + --------- — (-2 -.05 + y1) = 1.19512
2
are the first two steps. In this case, the equation
specifying yn+i is somewhat more complicated when
yn = 3 + tn - yn was substituted. When designing the steps to
calculate yn+1 on a computer, yn can be calculated first and
thus the simpler formula for yn+1 can be used. The exact
solution is y(t) = 2 + t - e-t, so y(.1) = 1.19516,
y(.2) = 1.38127, y(.3) = 1.55918 and y(.04) = 1.72968, so the
approximations using h = .0125 are quite accurate to five
decimal places.
4. In this case yn = 2tn + e"tnYn and thus the improved Euler
formula is
[(2tn + e"tnYn) + 2tn+1 + e-tn+l(yn + hyn) ] h
yn+1 = yn + -------------------------------------------------. For
2
n = 0, 1, 2 we get y1 = 1.05122, y2 = 1.10483 and y3 = 1.16072 for h = .05.
10. See Problem 4.
11. The improved Euler formula is
f(tn,yn) + f(tn+^ yn +hf(tn,yn)) ,
yn+1 = yn + ---------------------------------------- h. As suggested
2
in the text, it's best to perform the following steps when
n
n
n
164
Section 8.2
implementing this formula: let k1 = (4 - tnyn)/(1 + y^),
k2 = yn + hk1 and k3 = (4 - tn+1k2)/(1 + k2). Then
yn+1 = yn + (k1 + k3)h/2.
14a. Since f (tn + h) = f(tn+1) we have, using the first part of
Eq.(5) and the given equation,
en+1 = f (tn+1) - yn+1 = [f (tn)-yn] + [f (tn) -
yn + f(tn+h, yn+hy' n) _ - ,
] h + f"(tn)h2/2! + f"'(tn)h3/3!.
2
Since yn = f(tn) and y^ = f'(tn) = f(tn,yn) this reduces to
en+1 = f"(tn)h2/2! - {f[tn+h, yn + hf(tn,yn)]
- f(tn,yn)}h/2! + f"'(t n)h3/3!, which can be written in the form of Eq.(i).
14b. First observe that y' = f(t,y) and y" = ft(t,y) +
fy(t,y)y'. Hence f"(tn) = ft(tn,yn) + fy(tn,yn)f(tn,yn). Using the given Taylor series, with a = tn, h = h, b = yn and k = hf(tn,yn) we have
f[tn+h,yn+hf(tn,yn)] = f(tn,yn)+ft(tn,yn)h+fy(tn,yn)hf(tn,yn)
:yy(X,h)h2f2(tn,
n
+ [ftt(X, h )h2+2fty(X, h )h2f(tn,yn)+fyy(X, h )h2f2 (tn,yn)]/2!
where tn < X < tn + h and |h-yn| < h|f(tn,yn)|.
Substituting this in Eq.(i) and using the earlier expression for f"(tn) we find that the first term on the right side of Eq.(i) reduces to
-[ftt(X,h) + 2fty(X,h)f(tn,yn) + fyy (X, h )f2 (tn,yn)]h3/4, which is proportional to h3 plus, possibly, higher order terms. The reason that there may be higher order terms is because X and h will, in general, depend upon h.
14c.If f(t,y) is linear in t and y, then ftt = fty = fyy = 0 and the terms appearing in the last formula of part (b) are all zero.
15. Since f(t) = [4t - 3 + 19exp(4t)]/16 we have
f'''(t) = 76exp(4t) and thus from Problem 14c, since f is linear in t and y, we find
en+i = 38[exp(4tn)]h3/3. Thus
|en+i| ? (38h3/3)exp(8) = 37,758.8h3 on 0 < t < 2. For n = 1, we have |e1| = |f(t1) - y1| < (38/3)exp(0.2)(.05)3 = .001934,
which is approximately 1/15 of the error indicated in Eq.(27) of the previous section.
Section 8.3
165
19. The Euler method gives
Yi = Yo + h(5t0 - 3j/yo ) = 2 + .1(-3^2) = 1.57574 and the improved Euler method gives
f(to,Yo) + f(ti,yi)
Yi = Yo + -------------------------- h
2
= 2 + [-3^2 + (.5 - 3V 1.57574 )].o5 = 1.62458.
Thus, the estimated error in using the Euler method is 1.62458 - 1.57574 = .o4884. Since we want our error tolerance to be no greater than .oo25 we need to adjust the step size downward by a factor of j/ .oo25/.o4884 @ .226. Thus a step size of h = (.1)(.23) = .o23 would be needed for the required local truncation error bound of .oo25.
24. The modified Euler formula is
Yn+1 = Yn + hf[tn + h/2, yn + (h/2)f(tn,yn)] where
f(t,y) = 5t - 3\fy . Thus
y 1 = 2 + .o5[5(to + .o25) - 3sqrt(2 + .o25(5to - 3j/2))]
= 1.79982 for to = o. The values obtained here are between the values for h = .o5 and for h = .o25 using the Euler method in Problem 2.
Section 8.3, Page 438
4. The Runge-Kutta formula is
Yn+1 = Yn + h(kn1 + 2kn2 + 2kn3 + kn4)/6 where kn1, kn2 etc. are given by Eqs.(3). Thus for
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