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

ISBN 0-471-31999-6

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10. Since we are assuming continuity, either convert the monthly

payment into an annual payment or convert the yearly

interest rate into a monthly interest rate for 240 months. Then proceed as in Prob. 9.

12a. Using Eq. (15) we have - S = 800(1) or

S (800t), S(0) 100,000. Using an integrating factor and

integration by parts (or using a D.E. solver) we get S(t) t c. Using the I.C. yields c . Substituting this value into S, setting S(t) 0, and solving numerically for t yields t 135.363 months.

16a. This problem can be done numerically using appropriate D.E. solver software. Analytically we have

(.1.2sint)dt by separating variables and thus y(t) cexp(.1t.2cost). y(0) 1 gives c , so

y(t) exp(.2.1t.2cost). Setting y 2 yields ln2 .2 .1 .2cos, which can be solved numerically to give

2.9632. If y(0) , then as above,

y(t) exp(.2.1t.2cost). Thus if we set y 2 we get the

Section 2.3

15

same numerical equation for and hence the doubling time has not changed.

18. From Eq.(26) we have 19 = 7 + (20-7) or

k = -ln = ln(13/12). Hence if (T) = 15 we get: 15 = 7 +

13. Solving for T yields

T = ln(8/13)/-ln(13/12) = ln(13/8)/ln(13/12) min.

19. Hint: let Q(t) be the quantity of carbon monoxide in the

room at any time t. Then the concentration is given by x(t) = Q(t)/1200.

20a. The required I.V.P. is dQ/dt = kr + P - r,

Q(0) = V. Since c = Q(t)/V, the I.V.P. may be rewritten Vc(t) = kr + P - rc, c(0) = , which has the solution c(t) = k + + ( - k - ).

20b. Set k = 0, P = 0, t = T and c(T) = .5 in the solution found

in (a).

21a.If we measure x positively upward from the ground, then

Eq.(4) of Section 1.1 becomes m = -mg, since there is no air resistance. Thus the I.V.P. for v(t) is dv/dt = -g, v(0) = 20. Hence = v(t) = 20 - gt and x(t) = 20t - (g/2) + c. Since x(0) =

30, c = 30 and x(t) = 20t - (g/2)t + 30. At the maximum

height v(tm) = 0 and thus

tm = 20/9.8 = 2.04 sec., which when substituted in the equation for x(t) yields the maximum height.

2

21b. At the ground x(tg) = 0 and thus 20tg - 4.9tg + 30 = 0.

dv 1

22. The I.V.P. in this case is m— = v - mg, v(0) = 20,

dt 30

where the positive direction is measured upward. dv

2 4a. The I.V.P. is m— = mg - .75v, v(0) = 0 and v is

dt

measured positively downward. Since m = 180/32, the D.E.

dv 2 _2t/i5

becomes — = 32 - v and thus v(t) = 240(1-e / ) so

dt 15

that v(10) = 176.7 ft/sec.

24b. Integration of v(t) as found in (a) yields

x(t) = 240t + 1800(e 2t/15-1) where x is measured positively down from the altitude of 5000 feet. Set

16

Sect ion

2 . 3

t = 10 to find the distance traveled when the parachute opens.

dv

24c. After the parachute opens the I.V.P. is m— = mg-12v,

dt

v(0) = 176.7, which has the solution

-32t/15

v(t) = 161.7e + 15 and where t = 0 now represents

the time the parachute opens. Letting t^<^> yields the limiting velocity.

24d. Integrate v(t) as found in (c) to find

x(t) = 15t - 75.8e-32t/15 + C2. C2 = 75.8 since x(0) = 0,

x now being measured from the point where the parachute opens. Setting x = 3925.5 will then yield the length of time the skydiver is in the air after the parachute opens.

26a. Again, if x is measured positively upward, then Eq.(4) of

dv

Sect.1.1 becomes m— = -mg - kv.

dt

mg mg -kt/m

26b.From part (a) v(t) = -— + [v0 + ]e / . As k ^ 0

k 0 k

this has the indeterminant form of -• + •. Thus rewrite

v(t) as v(t) = [-mg + (v0k + mg)e-kt/m ]/k which has the

indeterminant form of 0/0, as k ^ 0 and hence L'Hopital's Rule may be applied with k as the variable.

27a. The equation of motion is m(dv/dt) = w-R-B which, in this problem, is

4 3 4 3 4 3

— pa p(dv/dt) = “pa pg - 6ppav - “pa pg. The limiting

velocity occurs when dv/dt = 0.

27b. Since the droplet is motionless, v = dv/dt = 0, we have

4 3 4 3

the equation of motion 0 = (—)pa pg - Ee - (—)pa pg,

33

where p is the density of the oil and p' is the density

of air. Solving for e yields the answer.

28. All three parts can be answered from one solution if k

represents the resistance and if the method of solution

of Example 4 is used. Thus we have dv dv

m— = mv = mg - kv, v(0) = 0, where we have assumed

dt dx

the velocity is a function of x. The solution of this

Section 2.3

17

I.V.P. involves a logarithmic term, and thus the answers to parts (a) and (c) must be found using a numerical procedure.

22 29b. Note that 32 ft/sec = 78,545 m/hr .

30. This problem is the same as Example 4 through Eq.(29).

V gR

In this case the I.C. is v(XR) = vo, so c =

2 1+X '

2gR

The escape velocity,ve, is found by noting that vo >

1+X

2

in order for v to always be positive. From Example 4, the escape velocity for a surface launch is ve(0) = V"2gR . We want the escape velocity of xo = XR to have the relation ve(^R) = .85ve(0), which yields

X = (0.85)-2 - 1 @ 0.384. If R = 4000 miles then xo = XR = 1536 miles.

dx

31b. From part a) — = v = ucosA and hence dt

x(t) = (ucosA)t + d1. Since x(0) = 0, we have d1 = 0 and

dy

x(t) = (ucosA)t. Likewise = —gt + usinA and

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