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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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Section 2.7, Page 103
1d. The exact solution to this I.V.P. is y = f(t) = t + 2 -e-t.
3a. The Euler formula is yn+1 = yn + h(2yn - tn + 1/2) for n = 0,1,2,3 and with t0 = 0 and y0 = 1. Thus y1 = y0 + .1(2y0 - t0 + 1/2) = 1.25,
y2 = 1.25 + .1[2(1.25) - (.1) + 1/2] = 1.54,
y 3 = 1.54 + .1[2(1.54) - (.2) + 1/2] = 1.878, and
y4 = 1.878 + .1[2(1.878) - (.3) + 1/2] = 2.2736.
Section 2.7
27
3b.
3c.
3d.
4d.
6.
Use the same formula as in Problem 3a, except now h = .05 and n = 0,1...7. Notice that only results for n = 1,3,5 and 7 are needed to compare with part a.
Again, use the same formula as above with h = .025 and n = 0,1...15. Notice that only results for n = 3,7,11 and 15 are needed to compare with parts a and b.
y' = 1/2 - t + 2y is a first order linear D.E. Rewrite the equation in the form y' - 2y = 1/2 - t and multiply both sides by the integrating factor e-2t to obtain (e-2ty)' = (1/2 - t)e-2t. Integrating the right side by
parts and multiplying by e2t we obtain y = ce-2t + t/2.
The I.C. y(0) = 1 ? c = 1 and hence the solution of the I.V.P. is y = f(x) = e2t + t/2. Thus f(0.1) = 1.2714, f(0.2) = 1.59182, f(0.3) = 1.97212, and f(0.4) = 2.42554.
The exact solution to this I.V.P. is y = f(t) = (6cost + 3sint - 6e-2t)/5.
For y(0) > 0 the
solutions appear to converge. For y(0)<0 the solutions diverge.
All solutions seem to diverge.
28
Section
2 . 7
13a. The Euler formula is
4-tnyn
yn+1 = yn + h(--------), where t0 = 0 and
1 + yn
y0 = y(0) = -2. Thus, for h = .1, we get
y i = -2 + .1(4/5) = -1.92
4-.1(-1.92)
y2 = -1.92 + .1( ) = -1.83055
1 + (1.92)2
4-.2(-1.83055)
y3 = -1.83055 + .1(--------------------) = -1.7302
1+(1.83055)2
4-.3(-1.7302)
y4 = -1.7302 + .1(------------) = -1.617043
1 + (1.7302)2
4 - .4(-1.617043)
y5 = -1.617043 + .1(----------------------) = -1.488494.
1 + (1.617043)2
Thus, y(.5) @ -1.488494.
15a. The Euler formula is
3tn
yn+1 = yn + .1 ( ) , where t0 = 1 and y0 = 0. Thus
3y2 -4
3
y1 = 0 + .1 (-----) = -.075 and
-4
3(1.1) 2
y2 = -.075 + .1 ( ) = .166134.
3(.075) 2-4
15c. There are two factors that explain the large differences. From the limit, the slope of y, y', becomes very "large" for values of y near -1.155. Also, the slope changes sign at y = -1.155. Thus for part a,
y(1.7) @ y7 = -1.178, which is close to -1.155 and the slope y' here is large and positive, creating the large change in y8 @ y(1.8). For part b, y(1.65) @ -1.125, resulting in a large negative slope, which yields y(1.70) @ -3.133. The slope at this point is now positive and the remainder of the solutions "grow" to -3.098 for the approcimation to y(1.8).
16. For the four step sizes given, the approximte values for y(.8) are 3.5078, 4.2013, 4.8004 and 5.3428. Thus, since these changes are still rather "large", it is hard to give an estimate other than y(.8) is at least 5.3428. By
using h = .005, .0025 and .001, we find further
approximate values of y(.8) to be 5.576, 5.707 and 5.790. Thus a better estimate now is for y(.8) to be between 5.8
and 6. No reliable estimate is obtainable for y(1), which is consistent with the direction field of Prob.9.
Section 2.8
29
18. It is helpful, in understanding this problem, to also . 2 calculate y (tn) = yn(.1 y n - tn). For a = 2.38 this term
remains positive and grows very large for tn > 2. On the
other hand, for a = 2.37 this term decreases and
eventually becomes negative for tn @ 1.6 (for h = .01).
For a = 2.37 and h = .1, .05 and .01, y(2.00) has the
approximations of 4.48, 4.01 and 3.50 respectively. A
small step size must be used, due to the sensitivety of
the slope field, given by yn (.1yn - tn).
22. Using Eq.(8) we have yn+i = yn + h(2yn -1) = (1+2h)yn - h.
Setting n + 1 = k (and hence n = k-1) this becomes yk = (1 + 2h)yk-1 - h, for k = 1,2,... . Since y0 = 1,
we have y1 = 1 + 2h - h = 1 + h = (1 + 2h)/2 + 1/2, and hence y2 = (1 + 2h)y1 - h = (1 + 2h)2/2 + (1 + 2h)/2 - h = (1 + 2h)2/2 + 1/2; y3 = (1 + 2h)y2 - h = (1 + 2h) 3/2 + (1 + 2h)/2 - h
= (1 + 2h)3/2 + 1/2. Continuing in this fashion (or
using induction) we obtain yk = (1 + 2h)k/2 + 1/2. For fixed x > 0 choose h = x/k. Then substitute for h in the last formula to obtain yk = (1 + 2x/k)k/2 + 1/2. Letting k ^ • we find (See hint for Problem 20d.)
y(x) = yk ^ e2x/2 + 1/2, which is the exact solution.
Section 2.8, Page 113
1. Let s = t-1 and w(s) = y(t(s)) - 2, then when t = 1 and
y = 2 we have s = 0 and w(0) = 0. Also,
dw dw dt d dt dy
— =---------= —(y-2)— = — and hence
ds dt ds dt ds dt
dw 2 2
— = (s+1) + (w+2) , upon substitution into the given ds
D.E.
4a. Following Ex. 1 of the text, from Eq.(7) we have
fn+1(t) = I tf(s,f(s))ds, where f(t,f) = -1 - f for this
0
problem. Thus if f0(t) = 0, then f1(t) = -I tds = -t;
0
ft t2
f2(t) = -I (1-s)ds = -t + ;
2 0 2
30
Section
2 . 8
2 t2 .3
its t t
f 3(t) = -| (1-s + )ds = -t + - ;
3 J0 2 2 2-3
i
2
s
3
s
t
2
f4(t) = -| (1 - s + — - —)ds = -t + — - — + —.
3!
3!
4!
Based upon these we hypothesize that: fn(t) = X,
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