7
2 Chapter 1
K 1 K 2
0
2
4
6
8
10
12
Figure 7K: The graph of y (t) in Problem 32.
1 0
1
2
3
4
5
6
7
Figure 7L: The graph of g (t) in Problem 42.
2
1
0 2.5 5.0 7.5
K
1
Figure 7M: The graph of y (t) in Problem 22.
1.0
0.5
0 2 4 6 8 10 12
K K
0.5
1.0
Figure 7N: The graph of y (t) in Problem 24.
266
CHAPTER 8: Series Solutions of Dierential Equations
EXERCISES 8.1: Introduction: The Taylor Polynomial Approximation
2. 2 + 4x + 8x2 + 4. 1 2 x2 + 1 6 1 6 x3 1 120 1 20 x5 +
6. x 8. 1
x3 +
x5 + x4 +
sin 1 2
x2 +
(cos 1)(sin 1) 24
10. (a) p3 (x) =
1x ++ 24 1 24 (b) 3 = 4! (2 )5 (c)
x2 x3 + 8 16 4 2 1 11 = 5 0.00823 5 24 2 (3/2) 3
2 1 1 p3 = 0.00260 3 2 384 (d) See Fig. 8A on page 276 12. The dierential equation implies that the functions y (x), y (x), and y (x) exist and continuous. Furthermore, y (x) can be obtained by dierentiating the other terms: y = py p y qy q y + g . Since p, q , and g have derivatives of all orders, subsequent dierentiations display the fact that, in turn, y , y (4) , y (5) , etc. all exist. 14. (a) t + t2 t3 + 2 6 (b) For r = 1, y (t) = 1 (1/2)t2 4t4 + ; For r = 1, y (t) = 1 + (1/2)t2 (49/12)t4 + (c) For small t in part (b), the hard spring recoils but the soft spring extends. 16. 1 t2 t4 + + 2 4 267
Chapter 8
EXERCISES 8.2: 2. (, ) 4. [2, 4] 6. (3, 1) 8. (a) (b) 11 , 22 Power Series and Analytic Functions
11 , 22 (c) (, )
(d) (, ) (e) (, ) (f )
11 , 22 (n + 2)2 2n+3 + (x 1)n (n + 3)! 2n+1 2 3 x3 + 2 15 x5 +
10.
n=0
12. x 14. 1 16. 1
1 2
x+
1 4
x2
1 24
x3 +
18.
k=0
(1)k x 2k = cos x (2k )! n(n 1)an xn2
20.
n=2
22.
k=0
(1)k x3 x5 2k+1 x =x + (2k + 1)(2k + 1)! 18 600 (k 2)(k 3)ak2 xk
24.
k=4
26.
k=4
ak 3 k x k
268
Exercises 8.3
30.
n=0
(1)n (x 1)n (1)n+1 n x n 1 2 (x 1) 1 8 (x 1)2 + 1 16 (x 1)3 5 128 (x 1)4 +
32.
n=1
34. 1 +
36. (a) Always true (b) Sometimes false (c) Always true (d) Always true
38.
n=0
x4 x6 (1)n 2n+2 x = x2 + n+1 2 3 Power Series Solutions to Linear Dierential Equations
EXERCISES 8.3: 2. 0 4. 1, 0 6. 1
8. No singular points 10. x 1 and x = 2 12. y = a0 x2 x3 1+x+ + + 2! 3!
= a0
n=0
xn = a0 ex n!
14. a0 1 16. a0 1 18. a0 1 +
x2 x3 + + a1 x + 2 6 x2 x3 x3 + + a1 x + x2 + + 2 3 2 x2 x3 + + a1 x + + 6 27 x 2k x2k+1 + a1 (1)k = a0 cos x + a1 sin x (2k )! (2k + 1)! k=0 269
= a0 ex + (a1 a0 ) xex
20. a0
k=0
(1)k
Chapter 8
22. a3k+2 = 0, k = 0, 1, . . . 1 4 (3k 5)(3k 2) 3k x a0 1 + (1)k (3k )! k=1
+a1
x+
k=1
(1)k
2 5 (3k 4)(3k 1) 3k+1 x (3k + 1)!
24. a0
(1)k (2k 3)2 (2k 5)2 32 2k x2 x4 + + x + a1 x 1 2 24 k=3 (2k )! x2 x3 x4 + + + 2 2 3 x2 x3 x4 + 2 3 8 5x2 x3 2 6
26. x +
28. 1
30. 1 + x +
32. (a) If a1 = 0, then y (x) is an even function; (d) a0 = 0, a1 > 0 36. 3 9t2 + t3 + t4 + 2 Equations with Analytic Coecients
EXERCISES 8.4: 2. Innite 4. 2 6. 1
8. a0 1 2(x + 1) + 3(x + 1)2
10 (x + 1)3 + 3
1 1 1 1 10. a0 1 (x 2)2 (x 2)3 + + a1 (x 2) + (x 2)2 (x 2)3 + 4 24 4 12 1 2 7 12. a0 1 + (x + 1)2 + (x + 1)3 + + a1 (x + 1) + 2(x + 1)2 + (x + 1)3 + 2 3 3 14. 1 + x + x2 + 16. t + 270 5x3 + 6
t3 t4 t5 + + + 3 12 24