Solução_Calculo_Stewart_6e

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F. dy dt =2− 3y dy . Rewriting this equation as 100 + 2t I(t) =exp TX.10 SECTION 9.6 PREDATOR-PREY SYSTEMS ¤ 399 dt + 3 y =2, we see that it is linear. 100 + 2t =exp 3 ln(100 + 2t) = (100 + 2t) 3/2 2 3 dt 100 + 2t Multiplying the differential equation by I(t) gives (100 + 2t) 3/2 dy dt +3(100+2t)1/2 y =2(100+2t) 3/2 [(100 + 2t) 3/2 y] 0 =2(100+2t) 3/2 ⇒ (100 + 2t) 3/2 y = 2 5 (100 + 2t)5/2 + C ⇒ y = 2 (100 + 2t)+C(100 + 5 2t)−3/2 .Now0=y(0) = 2 (100) + C · 5 100−3/2 =40+ 1 C ⇒ C = −40,000,so 1000 2 y = (100 + 2t) − 40,000(100 + 5 2t)−3/2 kg. From this solution (no pun intended), we calculate the salt concentration at time t to be C(t) = y(t) 100 + 2t = −40,000 (100 + 2t) + 2 kg −40,000 . In particular, C(20) = 5/2 5 L 140 + 2 5/2 5 ≈ 0.2275 kg L and y(20) = 2 5 (140) − 40,000(140)−3/2 ≈ 31.85 kg. 35. (a) dv dt + c m v = g and I(t) =e (c/m) dt = e (c/m)t , and multiplying the differential equation by I(t) gives e (c/m)t dv dt + vce(c/m)t 0 = ge (c/m)t ⇒ e v (c/m)t = ge (c/m)t . Hence, m v(t) =e −(c/m)t ge (c/m)t dt + K = mg/c + Ke −(c/m)t . But the object is dropped from rest, so v(0) = 0 and K = −mg/c. Thus, the velocity at time t is v(t) =(mg/c)[1 − e −(c/m)t ]. (b) lim t→∞ v(t) =mg/c (c) s(t) = v(t) dt =(mg/c)[t +(m/c)e −(c/m)t ]+c 1 where c 1 = s(0) − m 2 g/c 2 . s(0) is the initial position, so s(0) = 0 and s(t) =(mg/c)[t +(m/c)e −(c/m)t ] − m 2 g/c 2 . ⇒ 9.6 Predator-Prey Systems 1. (a) dx/dt = −0.05x +0.0001xy. Ify =0,wehavedx/dt = −0.05x, which indicates that in the absence of y, x declines at a rate proportional to itself. So x represents the predator population and y represents the prey population. The growth of the prey population, 0.1y (from dy/dt =0.1y − 0.005xy), is restricted only by encounters with predators (the term −0.005xy). The predator population increases only through the term 0.0001xy; that is, by encounters with the prey and not through additional food sources. (b) dy/dt = −0.015y +0.00008xy. Ifx =0,wehavedy/dt = −0.015y, which indicates that in the absence of x, y would decline at a rate proportional to itself. So y represents the predator population and x represents the prey population. The growth of the prey population, 0.2x (from dx/dt =0.2x − 0.0002x 2 − 0.006xy =0.2x(1 − 0.001x) − 0.006xy), is restricted by a carrying capacity of 1000 [from the term 1 − 0.001x =1− x/1000] and by encounters with predators (the term −0.006xy). The predator population increases only through the term 0.00008xy;thatis,byencounterswiththeprey and not through additional food sources.
F. 400 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS TX.10 3. (a) At t =0, there are about 300 rabbits and 100 foxes. At t = t 1, the number of foxes reaches a minimum of about 20 while the number of rabbits is about 1000. Att = t 2 , the number of rabbits reaches a maximum of about 2400, while the number of foxes rebounds to 100. Att = t 3, the number of rabbits decreases to about 1000 and the number of foxes reaches a maximum of about 315. Ast increases, the number of foxes decreases greatly to 100, and the number of rabbits decreases to 300 (the initial populations), and the cycle starts again. (b) 5. 7. dW dR = −0.02W +0.00002RW 0.08R − 0.001RW 0.08 − 0.001W W dW = −0.02 + 0.00002R R ⇔ (0.08 − 0.001W )RdW =(−0.02 + 0.00002R)W dR ⇔ dR ⇔ 0.08 W − 0.001 dW = − 0.02 R +0.00002 dR ⇔ 0.08 ln|W | − 0.001W = −0.02 ln|R| +0.00002R + K ⇔ 0.08 ln W +0.02 ln R =0.001W +0.00002R + K ⇔ ln W 0.08 R 0.02 =0.00002R +0.001W + K ⇔ W 0.08 R 0.02 = e 0.00002R+0.001W +K ⇔ R 0.02 W 0.08 = Ce 0.00002R e 0.001W ⇔ R 0.02 W 0.08 e 0.00002R e 0.001W dy −ry + bxy = C. In general, if = dx kx − axy ,thenC = xr y k e bx e ay . 9. (a) Letting W =0gives us dR/dt =0.08R(1 − 0.0002R). dR/dt =0 ⇔ R =0or 5000. SincedR/dt > 0 for 0
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