Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 9 REVIEW ¤ 405
(b) x and y are constant ⇒ x 0 =0and y 0 =0 ⇒
0=0.4x(1 − 0.000005x) − 0.002xy
0=−0.2y +0.000008xy
⇒
0=0.4x[(1 − 0.000005x) − 0.005y]
0=y(−0.2+0.000008x)
The second equation is true if y =0or x = 0.2
0.000008
=25,000. Ify =0in the first equation, then either x =0
or x =
1
0.000005
=200,000. Ifx =25,000, then0=0.4(25,000)[(1 − 0.000005 · 25,000) − 0.005y] ⇒
0=10,000[(1 − 0.125) − 0.005y] ⇒ 0 = 8750 − 50y ⇒ y =175.
Case (i):
y =0, x =0: Zero populations
Case (ii): y =0, x = 200,000: In the absence of birds, the insect population is always 200,000.
Case (iii):
x =25,000, y =175: The predator/prey interaction balances and the populations are stable.
(c) The populations of the birds and insects fluctuate
(d)
around 175 and 25,000, respectively, and
eventually stabilize at those values.
25. (a) d 2 y
dx 2 = k
1+
2 dy
.
dx
Setting z = dy dz
,weget
dx dx = k √ 1+z 2
⇒
dz
√ = kdx. Using Formula 25 gives
1+z
2
ln z + √ 1+z 2 = kx + c ⇒ z + √ 1+z 2 = Ce kx [where C = e c ] ⇒ √ 1+z 2 = Ce kx − z ⇒
1+z 2 = C 2 e 2kx − 2Ce kx z + z 2 ⇒ 2Ce kx z = C 2 e 2kx − 1 ⇒ z = C 2 ekx − 1
2C e−kx .Now
dy
dx = C 2 ekx − 1
2C e−kx ⇒ y = C 2k ekx + 1
2Ck e−kx + C 0 . From the diagram in the text, we see that y(0) = a
and y(±b) =h. a = y(0) = C 2k + 1
2Ck + C 0 ⇒ C 0 = a − C 2k − 1
2Ck
⇒
y = C 2k (ekx − 1) + 1
2Ck (e−kx − 1) + a. Fromh = y(±b), wefind h = C 2k (ekb − 1) + 1
2Ck (e−kb − 1) + a
and h = C 2k (e−kb − 1) + 1
2Ck (ekb − 1) + a. Subtracting the second equation from the first, we get
0= C e kb − e −kb
k 2
− 1 e kb − e −kb
= 1 Ck 2 k
C − 1
sinh kb.
C
Now k>0 and b>0, sosinh kb > 0 and C = ±1. IfC =1,then
y = 1
2k (ekx − 1) + 1
2k (e−kx − 1) + a = 1 e kx + e −kx
− 1 k 2 k + a = a + 1 (cosh kx − 1). IfC = −1,
k
then y = − 1
2k (ekx − 1) − 1
2k (e−kx − 1) + a = −1
k
e kx + e −kx
+ 1 2 k + a = a − 1 (cosh kx − 1).
k
Since k>0, cosh kx ≥ 1,andy ≥ a,weconcludethatC =1and y = a + 1 (cosh kx − 1),where
k