Solução_Calculo_Stewart_6e

F.
404 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
TX.10
15. (a) Using (4) and (7) in Section 9.4, we see that for dP
dt =0.1P 1 − P
with P (0) = 100, wehavek =0.1,
2000
K =2000, P 0 =100,andA =
P (t) =
2000 − 100
100
2000
2000
and P (20) = ≈ 560.
1+19e−0.1t 1+19e−2 2000
(b) P = 1200 ⇔ 1200 =
⇔ 1+19e −0.1t = 2000
1+19e −0.1t 1200
e −0.1t =
2
3 /19 ⇔ −0.1t =ln
2
⇔ t = −10 ln 2 ≈ 33.5.
57 57
17. (a) dL
dt ∝ L ∞ − L ⇒ dL
dt = k(L ∞ − L)
⇒
=19. Thus, the solution of the initial-value problem is
dL
L ∞ − L =
⇔ 19e −0.1t = 5 3 − 1 ⇔
kdt ⇒ −ln |L ∞ − L| = kt + C ⇒
ln |L ∞ − L| = −kt − C ⇒ |L ∞ − L| = e −kt−C ⇒ L ∞ − L = Ae −kt ⇒ L = L ∞ − Ae −kt .
At t =0, L = L(0) = L ∞ − A ⇒ A = L ∞ − L(0) ⇒ L(t) =L ∞ − [L ∞ − L(0)]e −kt .
(b) L ∞ =53cm, L(0) = 10 cm, and k =0.2 ⇒ L(t) =53− (53 − 10)e −0.2t =53− 43e −0.2t .
19. Let P represent the population and I the number of infected people. The rate of spread dI/dt is jointly proportional to I and
to P − I, so for some constant k, dI
1
dt = kI(P − I) =(kP)I − I
. From Equation 9.4.7 with K = P and k replaced by
P
21.
kP,wehaveI(t) =
P
1+Ae = I 0 P
−kP t I 0 +(P − I 0 )e . −kP t
Now, measuring t in days, we substitute t =7, P = 5000, I 0 = 160 and I(7) = 1200 to find k:
1200 =
e −35,000k =
160 · 5000
2000
⇔ 3=
⇔ 480 + 14,520e −35,000k =2000 ⇔
160 + (5000 − 160)e −5000·7·k 160 + 4840e −35,000k
2000 − 480
14,520
⇔ −35,000k =ln 38
363
I =5000× 80% = 4000,andsolvefort: 4000 =
160 + 4840e −5000kt =200 ⇔ e −5000kt =
t =
−1
5000k ln 1
121 = 1
1
to be infected.
dh
dt = − R h
V k + h
38
ln
7 363
⇒
· ln 1
121
k + h
h
=7· ln 121
ln 363
38
dh =
⇔ k = −1 38
ln ≈ 0.00006448. Next, let
35,000 363
160 · 5000
200
⇔ 1=
⇔
160 + (5000 − 160)e −k·5000·t 160 + 4840e −5000kt
200 − 160
4840
− R
dt
V
⇔ −5000kt =ln 1
121
≈ 14.875. So it takes about 15 days for 80% of the population
⇒
1+ k
dh = − R h V
⇔
1 dt ⇒
h + k ln h = − R V
t + C. This equation gives a relationship between h and t, but it is not possible to isolate h and express it in
terms of t.
23. (a) dx/dt =0.4x(1 − 0.000005x) − 0.002xy, dy/dt = −0.2y +0.000008xy. Ify =0,then
dx/dt =0.4x(1 − 0.000005x),sodx/dt =0 ⇔ x =0or x = 200,000, which shows that the insect population
increases logistically with a carrying capacity of 200,000. Sincedx/dt > 0 for 0