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

ISBN 0-471-31999-6

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1. Linear 2. Homogeneous

3. Exact 4. Linear equation in x(y)

5. Exact 6. Linear

2 dy dy

7. Letting u = x yields --------- = 2x— and thus

dx du

du 3

— - 2yu = 2y which is linear in u(y). dy

34 Miscellaneous Problems

8. Linear 9. Exact

10. Integrating factor depends on x only 11. Exact

12. Linear 13. Homogeneous

14. Exact or homogeneous 15. Separable

16. Homogeneous 17. Linear

18. Linear or homogeneous 19. Integrating factor depends on x only

20. Separable 21. Homogeneous

22. Separable 23. Bernoulli equation

24. Separable 25. Exact

26. Integrating factor depends on x only 27. Integrating factor depends on x only

28. Exact 29. Homogeneous

30. Linear equation in x(y) 31. Separable

32. Integrating factor depends on y only.

35

CHAPTER 3

Section 3.1, Page 136

-TX rt T ? T / rt n 2 rt

3. Assume y = e , which gives y = re and y = re .

2 rt

Substitution into the D.E. yields (6r -r-1)e = 0.

rt

Since e no, we have the characteristic equation 6r2-r-1 = 0, or (3r+1)(2r-1) = 0. Thus

r = -1/3, 1/2 and y = c1et/2 + c2e-t/3.

2

5. The characteristic equation is r + 5r = 0, so the roots

are r1 = 0, and r2 = -5. Thus

0t -5t -5t

y = c1e + c2e = c1 + c2e .

2

7. The characteristic equation is r - 9r + 9 = 0 so that

r = (9±\j 81-36 )/2 = (9±3^/5)/2 using the quadratic

formula. Hence

y = c1exp[(9 + 3^/5 )t/2] + c2exp[(9-3^/5 )t/2].

rt

10. Substituting y = e in the D.E.

we obtain the characteristic

2

equation r + 4r + 3 = 0, which has the roots r1 = -1, r2 = -3.

-t -3t

Thus y = c1e + c2e and

-t -3t

y = -c1e - 3c2e .

Substituting t = 0 we then have c1 + c2 = 2 and -c1 - 3c2 = -1,

yielding c1 = 5/2 and

5 -t 1 -3t

c2 = -1/2. Thus y = —e - —e

2 2 2

and hence y ^ 0 as t ^ •.

2

15. The characteristic equation is r + 8r - 9 = 0, so that r1 = 1 and r2 = -9 and the general solution is

t -9t

y = c1e + c2e . Since the I.C. are given at t = 1, it

is convenient to write the general solution in the form

(t-1) -9(t-1)

y = k1e + k2e . Note that

-1 9

c1 = k1e and c2 = k2e . The advantage of the latter form of the general solution becomes clear when we apply

36

Section 3 . 1

is y(t ) = ce + ce .

1 2

3

c - c = - — , yielding

1 2 4

(t) = 1t - t

— e - e = 0 or

4

the I.C. y(1) = 1 and y/(1) = 0. This latter form of y

(t-1) -9(t-1)

gives y = k1e - 9k2e and thus setting t = 1 in

y and y' yields the equations k2 + k2 = 1 and

ki - 9k2 = 0. Solving for k2 and k2 we find that

(t-1) -9(t-1) (t-1)

y = (9e + e )/10. Since e has a positive

exponent for t > 1, y ^ • as t ^ • .

17. Comparing the given solution to Eq(17), we see that r = 2

1

and r = -3 are the two roots of the characteristic 2

2

equation. Thus we have (r-2)(r+3) = 0, or r + r - 6 = 0

as the characteristic equation. Hence the given solution

is for the D.E. y" + y' - 6y = 0.

19. The roots of the characteristic equation are r = 1, -1

t -t

5

y(0) = c + c = — and y'(0

124

1 t -t

y(t) = —e + e . From this

4

2t

e = 4 or t = ln2. The second derivative test or a graph of the solution indicates this is a minimum point.

-t 2t

21. The general solution is y = c1e + c2e . Using the I.C.

we obtain c1 + c2 = a and -c1 + 2c2 = 2, so adding the two

equations we find 3c2 = a + 2. If y is to approach zero

as t —— •, c2 must be zero. Thus a = -2.

24. The roots of the characteristic equation are given by

-2t (a-1)t

r = -2, a - 1 and thus y(t) = ce + ce . Hence,

12

for a < 1, all solutions tend to zero as t — •. For a > 1, the second term becomes unbounded, but not the first, so there are no values of a for which all solutions become unbounded.

2

25a. The characteristic equation is 2r + 3r - 2 = 0, so

r1 = -2 and r2 = 1/2 and y = c1e 2t + c2et/2. The I.C.

1

yield c2 + c2 = 2 and -2c2 + —c2 = -p so that c2 = (1 + 2b)/5 and c2 = (4-2b)/5.

Section 3.1

37

25c.

27.

28.

30.

34.

37.

-2t

From part (a), if b = 2 then y(t) = e and the solution simply decays to zero. For b > 2, the solution becomes unbounded negatively, and again there is no minimum point.

-t

The second solution must decay faster than e , so choose

-2t -3t

e or e etc. as the second solution. Then proceed as in Problem 17.

Let v = y', then v' = y" and thus the D.E. becomes 22 tv' + 2tv - 1 = 0 or tv' + 2tv = 1. The left side is 2

recognized as (t v)' and thus we may integrate to obtain 2

tv = t + c (otherwise, divide both sides of the D.E. by 22 t and find the integrating factor, which is just t in

this case). Solving for v = dy/dt we find

2

dy/dt = 1/t + c/t so that y = lnt + c /t + c .

1 2

Set v = y', then v' = y" and thus the D.E. becomes

v' + tv2 = 0. This equation is separable and has the -1 2 2

solution -v + t /2 = c or v = y' = -2/(c1 - t ) where

c1 = 2c. We must consider separately the cases c1 = 0,

2

c1 > 0 and c1 < 0. If c1 = 0, then y' = 2/t or

y = -2/t + c2. If c1 > 0, let c1 = k . Then

y' = -2/(k-t ) = -(1/k)[1/(k-t) + 1/(k+t)], so that

y = (1/k)ln|(k-t)/(k+t)l+c2. If c1 < 0, let c1 = -k2.

2 2 -1 Then y' = 2/(k + t ) so that y = (2/k)tan (t/k) + c2.

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