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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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14. To verify that the given vector is the general solution of the corresponding system, it is sufficient to substitute it into the D.E. Note also that the two terms in x(c) are linearly independent. If we seek a solution of the form x = ?(t)u(t) then we find that the equation corresponding to Eq.(26) is t?(t)u'(t) = g(t), where
?(t) =
't 1/t' 2 1-t2
and g(t) = . Thus
vt 3/t, v 2t /
u' = (1/t)?-1(t)g (t). Using a computer algebra system or row operations on ? and I, we find that
3
?
-1
/3/2t -1/2t ^
-1/2 t/2
_ 2
and hence u1 =
2t
31 - - and 2 2 t
. 1 t 3 3t
u2 = + + t, which yields u1 = - - lnt + c1
2 2 2t 2
1 t3 t2
and u2 = - t + + + c2. 2 6 2
Multiplication of u by ?(t) yields the desired solution.
160
CHAPTER 8
Section 8.1, Page 427
In the following problems that ask for a large number of numerical calculations the first few steps are shown. It is then necessary to use these samples as a model to format a computer program or calculator to find the remaining values.
1a. The Euler formulas is yn+1 = yn + h(3 + tn - yn) for
n = 0,1,2,3... and with t0 = 0 and y0 = 1. Thus
y1 = 1 + .05(3 + 0 - 1) = 1.1
y2 = 1.1 + .05(3 + .05 - 1.1) = 1.1975 @ y(.1)
y3 = 1.1975 + .05(3 + .1 - 1.1975) = 1.29263
y4 = 1.29263 + .05(3 + .15 - 1.29263) = 1.38549 @ y(.2).
1c. The backward Euler formula is yn+i = yn + h(3 + tn+i - yn+i).
Solving this for yn+1 we find yn+1= [yn + h(3 + tn+1)]/(1+n).
1 + .05(3.05)
Thus y1 = = 1.097619 and
1.05
1.097619 + .05(3.1)
y2 = -------------------------- = 1.192971.
2 1.05
y0 + 2t0y0 (.5)2 + 0
5a. y1 = y0 + h------------- = .5 + .05-------- = .504167
3 + t20 3 + 0
(.504167)2 + 2(.05)(.504167)
y2 = .504167 + .05----------------------------------- = .509239
3 + (.05) 2
y2 + 2(.05)y1
5c. y1 = .5 + .05----------------------------------, which is a quadratic
3 + (.05) 2
equation in y1. Using the quadratic formula, or an equation solver, we obtain y1 = .5050895. Thus
y2 + 2(.1)y2
y2 = .5050895 + .05-------------------- which is again quadratic
3 + (.1)2
in y2, yielding y2 = .5111273.
7a. For part a eighty steps must be taken, that is,
n = 0,1,...79 and for part b 160 steps must taken with
n = 0,1,...159. Thus use of a programmable calculator or
a computer is required.
7c. We have yn+1 = yn + h(.5-tn+1 + 2yn+1), which is linear in
yn + .5h -htn+1
yn+1 and thus we have yn+1 = ----------------------. Again, 80
1 - 2h
Section 8.1
161
steps are needed here and 160 steps in part d. In This case a spreadsheet is very useful. The first few, the middle three and last two lines are shown for h = .025:
n yn tn yn+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^ and y^ denote the approximate
values of f(tn), f'(tn), and f"(tn), respectively.
Keeping the first three terms in the Taylor series we have
yn+1 = yn + y'nh + yn h2/2
= yn + (1 - tn + 4yn)h + (3 - 4tn + 16yn)h2/2. For n = 0,
t0 = 0 and y0 = 1 we have
y i = 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 | < 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.
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