Page 832 - Advanced_Engineering_Mathematics o'neil

812 Answers to Selected Problems

1 1

y 2 (x) = c ∗ 1 − x + x 3
0 2
2 2 (3!)(3)(5)
1
4
+ x +···
4
2 (4!)(3)(5)(7)
(−1)

n
= c ∗ 1 + ∞ x n
0 n=1 n
2 n!(1 · 3···(2n − 1))

1 1 1
2 4 6 8
7. y 1 (x) = c 0 x + x + x + x +··· = c 0 x sinh(x),
3! 5! 7!
1 1 1

4
5
2
3
∗
y 2 (x) = c ∗ x − x + x − x + x −··· = c xe −x
0 0
2! 3! 4!

1
9. y 1 (x) = c 0 (1 − x), y 2 (x) = c ∗ 0 1 + (x − 1)ln((x − 2)/x)
2
CHAPTER FIVE APPROXIMATIONS OF SOLUTIONS
Section 5.1 Direction Fields
1. Figure A.8 3. Figure A.9 5. Figure A.10
Section 5.2 Euler’s Method
1. Approximate values are given in Table A.1.
3. See Table A.2.
In this case the exact solution y = 5e 3x 2 /2 increases so rapidly that the Euler approximations are increasingly
inaccurate as x increases from 0.
5. See Table A.3.
Section 5.3 Taylor and Modified Euler Methods
7. See Table A.4.
9. See Table A.5.
11. See Table A.6.

4
2

2
1
y (x) y(x)
0 0
–4 –2 0 2 4 –3 –2 –1 0 1 2 3
x x
–1
–2
–2
–4

FIGURE A.8 Direction field and solution FIGURE A.9 Direction field and solution
for Problem 1, Section 5.1 for Problem 3, Section 5.1

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October 14, 2010 17:50 THM/NEIL Page-812 27410_25_Ans_p801-866