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

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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/15

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 - 6pmav - 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(^R) = 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

dt

2

therefore y(t) = -gt /2 + (usinA)t + d2. Since y(0) = h we have d2 = h and y(t) = -gt /2 + (usinA)t + h.

31d. Let t,. be the time the ball reaches the wall. Then

w

L

x(tw) = L = (ucosA)tw and thus tw = --------------. For the ball

w w w ucosA

to clear the wall y(tw) > H and thus (setting L

tw = -------, g = 32 and h = 3 in y) we get

w ucosA

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