Page 823 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 803

7. (a) Since f (x, y) = 4 + y and ∂ f/∂y = 1 are continuous everywhere, the initial value problem has a unique solution
in some interval about 0.
(b) y =−4 + 7e x
(c) y 0 = 3,
x
y 1 = 3 + 7dt = 3 + 7x,
0
x 7

2
y 2 = 3 + (7 + 7t)dt = 3 + 7x + x ,
2
0
x 7 7 7

3
2
y 3 = 3 + 7 + 7t + t 2 dt = 3 + 7x + x + x ,
2 2 3!
0
x
7 2 7 3

y 4 = 3 + 7 + 7x + x + x dt
0 2 3!
7 7 7
2
4
3!
= 3 + 7x + x + x + x ,
2 3 4!
x

y 5 = 3 + (4 + y 4 (t))dt
0
7 7 7 7
2
4
5
3!
= 3 + 7x + x + x + x + x ,
2 3 4! 5!
x
y 6 = 3 + (4 + y 5 (t))dt
0
7 2 7 7 4 7 5 7 6
3!
= 3 + 7x + x + x + x + x + x .
2 3 4! 5! 6!
(d)
7 7 7
3
2
y n = 3 + 7x + x + x +··· + x n
2! 3! n!
Note that
n
1
y n =−4 + 7 x k
k!
k=0
x
so lim n→∞ y n (x) =−4 + 7e .
2
9. (a) Both f (x, y) = 2x and ∂ f/∂y = 0 are continuous for all (x, y).
3
(b) y = (2/3)x + 7/3
3
2
x
(c) y 0 ) = 3, y 1 = 3 + 2t dt = (2/3)x + 7/3
1
2
(d) Since f (x, y) = 2x is independent of y, y n (x) = y 1 (x) for all n ≥ 1 and the sequence of Picard iterates simply
repeats this term. Now
2 7 2
2
3
y = x + = 3 + 2(x − 1) + 2(x − 1) + (x − 1) 3
3 3 3
is the Taylor series of the solution about 1. For n ≥ 3, the nth partial sum of this expansion is the solution, so
y n → y and the Picard iterates converge to the solution.
CHAPTER TWO LINEAR SECOND-ORDER EQUATIONS
Section 2.1 The Linear Second-Order Equation
1
1. y(x) = c 1 sin(6x) + c 2 cos(6x); y(x) = sin(6x) − 5cos(6x)
3
−x
3. y(x) = c 1 e −2x + c 2 e ; y(x) = 4e −2x − 7e −x
x
x
x
x
5. y(x) = c 1 e cos(x) + c 2 e sin(x); y(x) = 6e cos(x) − 5e sin(x)
4x
7. y(x) = c 1 e + c 2 e −4x − x 2 + 1
4 2
3x
3x
9. y(x) = c 1 e cos(2x) + c 2 e sin(2x) − 8e x
Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s).
Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.
October 14, 2010 17:50 THM/NEIL Page-803 27410_25_Ans_p801-866