# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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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±3ä/"5)/2 using the quadratic

formula. Hence

y = c1exp[(9 + 3ä/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 / ë

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 + c e .

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 k1 + k2 = 1 and

k1 - 9k2 = 0. Solving for k1 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) = c e + c e . 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 c1 + c2 = 2 and -2c1 + —c2 = -p so that c1 = (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

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