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Math 211 Fall 2007: Solutions: HW #4 Instructor: S. Cautis 1. section 2.9, #8 Equilibrium points: y0 (t) = 0 and y1 (t) = 2 are equilibrium solutions. The solution y0 is stable and y1 is unstable. 2. section 2.9, #10 The equilibrium solutions are y0 (t) = −2. y1 (t) = − 21 , y2 (t) = 1 and y3 (t) = 2. The stable solutions are y0 and y2 . 3. section 2.9, #20 The equilibrium points are y0 = −3, y1 = −1 and y2 = 3, y1 is stable. 4. section 2.9, #28 The equilibrium points are x0 = −2, x1 = 0 and x2 = 1. x1 is stable. 5. section 3.1, #4 P (t) = Cert is our model. P (0) = 1000 also P (10) = 2 ∗ 1000 = 2000 = 1000e10r so, r = ln102 . ln 2 P (t0 ) = 10000 then 1000e 10 t0 = 10000 So t0 = 10 ∗ lnln10 2 . 6. section 3.1 #10 Find the maximum rate of growth for the logistic equation. The maximum rate of growth is the maximum of dP dt . To do this find the zeros of the second derivative which is: d2 P 2P P 0 0 = rP 0 (1 − k2 P ) dt2 = rP − r k This is zero when P = k2 or when the population is half of the maximum. 2 This is a maximum because of the sign of ddtP2 at P = k2 − > 0 and P = k2 + < 0. 7. section 3.1 #12 rt Kce P (t) = 1+ce rt P (0) = 1000, P (10) = 2 ∗ 1000 = 2000, K = 20, 000 Find P (25): 0 P (0) = 1000 = 20,000ce 1+ce0 1 c = 19 P (10) = 2000 = 1 r10 20,000∗ 19 e 1 r10 1+ 19 e 1 2000 = e10r 19 (20, 000 − 2000) Meaning, r = P (25) = 1 r25 20000∗ 19 e 1 r25 1+ 19 e is roughly 5083.75 1 10 ln( 19 9 ) 8. section 3.3 #4 M 0 = .0625M + D so (e−.0625t M )0 = e−.0625t D Which means 1 1 e−.0625t D + C or M (t) = − .0625 D + C(e.0625t ). e−.0625t M = − .0625 1 Now M (0) = 0 so C = .0625 D. We also know M (18) = 50, 000 50000 50000 C(e.0625∗18 ) − C = 50, 000 and C = e.0625∗18 −1 , D = .0625 e.0625∗18 −1 9. section 4.1 #10 k ∗ 34 = 5 ∗ 9.8 0 5y 00 + 20∗9.8 3 y = 0 and y(0) = .36 , y (0) = −.45 10. section 4.1 #14 (C1 cos(2t) + C2 sin(2t))00 + 4(C1 cos(2t) + C2 sin(2t)) = − 4C1 cos(2t) + −4C2 sin(2t) + 4(C1 cos(2t) + C2 sin(2t)) = 0