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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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= 3 + tn - was substituted. When designing the steps to
calculate +1 on a computer, can be calculated first and
thus the simpler formula for +1 can be used. The exact
solution is y(t) = 2 + t - e-t, so y(.i) = 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 = 2tn + e-tnyn and thus the improved Euler
formula is
[(2tn + e-tnyn) + 2tn+1 + e-tn+i(yn + hyp) ] h
+1 = + ------------------------------------------------- . For
2
n = 0, 1, 2 we get yi = 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+^ +hf(tn,yn)) ,
+1 = + --------------------------------------- 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 + ),
k2 = yn + hk1 and k3 = (4 - tn+1k2)/(1 + k2). Then
+1 = + (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) -
+ f(tn+h, yn+hy' ) _ - ,
-------------------------] h + f"(tn)h2/2! + f'''(tn)h3/3!.
2
Since yn = f(tn) and = f'(tn) = f(tn,yn) this reduces to
en+1 = f"(tn)h2/2! - {f[tn+h, + 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 jh-! < 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+1 = 38[exp(4tn)]h3/3. Thus
jen+1| < (38h3/3)exp(8) = 37,758.8h3 on 0 < t < 2. For n = 1, we have je1j = jf(t1) - y1j < (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
1 - 0 + h(5t0 - 3 ^/y0 ) - 2 + .1(-3^2) - 1.57574 and the improved Euler method gives
f(t0,y0) + f(t1,y1)
1 - 0 + ----------------------- h
2
- 2 + [-3^/2 + (.5 - ^1.57574 )].05 - 1.62458.
Thus, the estimated error in using the Euler method is 1.62458 - 1.57574 - .04884. Since we want our error tolerance to be no greater than .0025 we need to adjust the step size downward by a factor of V .0025/.04884 @ .226. Thus a step
size of h - (.1)(.23) - .023 would be needed for the required local truncation error bound of .0025.
24. The modified Euler formula is
+1 - + hf[tn + h/2, yn + (h/2)f(tn,yn)] where
f(t,y) - 5t - 3\fy . Thus
ӳ - 2 + .05[5(t0 + .025) - 3sqrt(2 + .025(5t0 - 3^/2))]
- 1.79982 for t0 - 0. The values obtained here are between the values for h - .05 and for h - .025 using the Euler method in Problem 2.
Section 8.3, Page 438
4. The Runge-Kutta formula is
+1 - + h(kn1 + 2kn2 + 2kn3 + kn4)/6 where kn^ kn2 etc. are given by Eqs.(3). Thus for
f(t,y) - 2t + e-ty, (t0,y0) - (0,1) and h - .1 we have k01 - 0 + e0 - 1
k02 - 2(0 + .05) + e-(0+.05)(1+.05k01) - 1.048854
k03 - 2 (.05) + e"(.05)(1+.05k2) - 1.048738
k04 - 2(.1) + e"(.1)(1+.1k3) - 1.095398 and hence
y(.1) @ 1 - 1 + .1(k01 + 2k02 + 2k03 + k04)/6 - 1.104843.
We have f(tn,yn) - (4 - tnyn)/(1 + ). Thus for t0 - 0,
0 - -2 and h - .1 we have
k01 - f(0,-2) - .8
k02 - f( .05, - 2 + .05(.8)) - f(.05, - 1.96) - . 846414,
3 - f( .05, - 2 + .05k02) - f(.05, - 1.957679) - .847983,
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