# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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Birkhoff, G., and Rota, G.-C., Ordinary Differential Equations (4th ed.) (New York: Wiley, 1989). Sagan, H., Boundary and Eigenvalue Problems in Mathematical Physics (New York: Wiley, 1961;

New York: Dover, 1989).

Weinberger, H., A First Course in Partial Differential Equations (New York: Wiley, 1965; New York: Dover, 1995).

Yosida, K., Lectures on Differential and Integral Equations (New York: Wiley-Interscience, 1960).

The following books are convenient sources of numerical and graphical data about Bessel and Legendre functions:

Abramowitz, M., and Stegun, I. A. (eds.), Handbook of Mathematical Functions (New York: Dover, 1965); originally published by the National Bureau of Standards, Washington, DC, 1964.

Jahnke, E., and Emde, F., Tables of Functions with Formulae and Curves (Leipzig: Teubner, 1938; New York: Dover, 1945).

The following books also contain much information about Sturm-Liouville problems:

Cole, R. H., Theory of Ordinary Differential Equations (New York: Irvington, 1968).

Hochstadt, H., Differential Equations: A Modern Approach (New York: Holt, 1964; New York: Dover, 1975).

Miller, R. K., and Michel, A. N., Ordinary Differential Equations (New York: Academic Press, 1982). Tricomi, F. G., Differential Equations (New York: Hafner, 1961).

Answers to Problems

CHAPTER 1 Section 1.1, page 8

1. y ^ 3/2 as t ^? 2. y diverges from 3/2 as t ^ ?

3. y diverges from -3/2 as t ^? 4. y 1/2 as t ^?

5. y diverges from 1/2 as t ^? 6. y diverges from 2 as t ^ ?

7. y = 3 y 8. y = 2 3y

See SSM for 9. y = y 2 10. y = 3y 1

detailed solutions to 11. y = 0 and y = 4 are equilibrium solutions; y ^ 4 if initial value is positive; y diverges

2, 4, 7, 11, 13, from 0 if initial value is negative.

15ab. 12. y = 0 and y = 5 are equilibrium solutions; y diverges from 5 if initial value is greater than 5; y ^ 0 if initial value is less than 5.

13. y = 0 is equilibrium solution; y ^ 0 if initial value is negative; y diverges from 0 if initial value is positive.

14. y = 0 and y = 2 are equilibrium solutions; y diverges from 0 if initial value is negative; y ^ 2 if initial value is between 0 and 2; y diverges from 2 if initial value is greater than

15. 2. (a) dq/dt = 300(102 q 106) (b) q ^ 104 g; no

16, 21, 22 16. dV/dt = kV2/3 for some k > 0.

17. (a) dq/dt = 500 0.4q (b) q ^ 1250 mg

18. (a) mv' = mg kv2 (b) v ^ ymg/ k (c) k = 2/49

19. y is asymptotic to t 3as t ^? 20. y ^ 0as t ^?

21. y ^ ?, 0, or ? depending on the initial value of y

22. y ^ ? or ? depending on the initial value of y

23. y ^ ? or ? or y oscillates depending on the initial value of y

679

680

Answers to Problems

See SSM for 24. y » or is asymptotic to V2t 1 depending on the initial value of y

detailed solution 25. y 0 and then fails to exist after some tf > 0

for 24 26. y or » depending on the initial value of y

Section 1.2, page 14

1b 1. (a) y = 5 + (yo 5)et (b) y = (5/2) + y (5/2)]e2t

(c) y = 5 + (yo 5)e2t

Equilibrium solution is y = 5 in (a) and (c), y = 5/2 in (b); solution approaches equilibrium faster in (b) and (c) than in (a).

2. (a) y = 5 + (yo 5)et (b) y = (5/2) + [yo (5/2)]e2t

(c) y = 5 + (yo 5)e2t

Equilibrium solution is y = 5 in (a) and (c), y = 5/2 in (b); solution diverges from equilibrium faster in (b) and (c) than in (a).

3. (a) y = ceat + (b/a)

(c) (i) Equilibrium is lower and is approached more rapidly. (ii) Equilibrium is higher.

(iii) Equilibrium remains the same and is approached more rapidly.

4ab, 6b, 4. (a) y(t) = ceat (b) y = ceat + (b/a)

8ab,10a, 13abc 5. y = Re^a + (b/a)

6. (a) T = 2ln18 = 5.78 months (b) T = 2ln[9oo/(9oo p0)] months

(c) po = 9oo(1 e6) = 897.8

7. (a) r = (ln2)/3o day1 (b) r = (ln2)/N day1

8. (a) T = 5 ln5o = 19.56 sec (b) 718.34 m

9. (a) dv/dt = 9.8, v(o) = o (b) T = V300/4.9 = 7.82 sec

(c) v = 76.68 m/sec

1o. (a) r = o.o2828 day1 (b) Q(t) = 100e0'02828t (c) T = 24.5 days

12. 162oln(4/3)/ ln2 = 672.4 years

13. (a) Q(t) = CV(1 eRC) (b) Q(t) CV = QL

(c) Q(t) = CVexp[ (t t1)/RC]

14. (a) Q = 3(1 1o4Q), Q(o) = o

(b) Q(t) = 1o4(1 e3t^1o ), t inhrs; after 1 year Q = 9277.77 g

(c) Q = -3 Q/1o4, Q(o) = 9277.77

(d) Q(t) = 9277.77e3(/1° , t inhrs; after 1 year Q = 67o.o7 g

(e) T = 2.6o years

15. (a) q = q/3oo, q(o) = 5ooo g (b) q(t) = 5000et/300 (c) no

(d) T = 300 ln(25/6) = 7.136 hr

(e) r = 250ln(25/6) = 256.78 gal/min

Section 1.3, page 22

2, 6, 8, 14, 16, 19 1. Second order, linear 2. Second order, nonlinear

3. Fourth order, linear 4. First order, nonlinear

5. Second order, nonlinear 6. Third order, linear

15. r =-2 16. r =±1

17. r = 2, 3 18. r = o, 1, 2

19. r = -1, 2 20. r = 1, 4

22, 26 21. Second order, linear 22. Second order, nonlinear

23. Fourth order, linear 24. Second order, nonlinear

CHAPTER 2 Section 2.1, page 38

1. (c) y = ce3t + (t/3) (1 /9) + e2t; y is asymptotic to t/3 1/9 as t » 1,2,3 2. (c) y = ce2t + t3e2t/3; y » as t »

3. (c) y = cet + 1 + t2et/2; y 1 as t »

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