Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 9 REVIEW ¤ 401
R = 1
0.0002
=5000 [as in part (a)]. If R = 1000, then0 = 1000[0.08(1 − 0.0002 · 1000) − 0.001W ] ⇔
0=80(1− 0.2) − W ⇔ W =64.
Case (i):
Case (ii):
Case (iii):
W =0, R =0: both populations are zero
W =0, R = 5000: seepart(a)
R =1000, W =64: the predator/prey interaction balances and the populations are stable.
(c) The populations of wolves and rabbits fluctuate around 64 and 1000, respectively, and eventually stabilize at those values.
(d)
9 Review
1. (a) A differential equation is an equation that contains an unknown function and one or more of its derivatives.
(b) The order of a differential equation is the order of the highest derivative that occurs in the equation.
(c) An initial condition is a condition of the form y(t 0 )=y 0 .
2. y 0 = x 2 + y 2 ≥ 0 for all x and y. y 0 =0only at the origin, so there is a horizontal tangent at (0, 0), but nowhere else. The
graph of the solution is increasing on every interval.
3. See the paragraph preceding Example 1 in Section 9.2.
4. See the paragraph next to Figure 14 in Section 9.2.
5. A separable equation is a first-order differential equation in which the expression for dy/dx can be factored as a function of x
times a function of y, thatis,dy/dx = g(x)f(y). We can solve the equation by integrating both sides of the equation
dy/f(y) =g(x)dx and solving for y.
6. A first-order linear differential equation is a differential equation that can be put in the form dy + P (x) y = Q(x),whereP
dx
and Q are continuous functions on a given interval. To solve such an equation, multiply it by the integrating factor
I(x) =e P (x)dx to put it in the form [I(x) y] 0 = I(x) Q(x) and then integrate both sides to get I(x) y = I(x) Q(x) dx,
that is, e P (x) dx y = e P (x) dx Q(x) dx. Solving for y gives us y = e − P (x) dx e P (x) dx Q(x) dx.