� 18.03 Problem Set 3: Part II Solutions Part I points: 8. 8, 9. 12, 11. 9, 12. 7. 8. (a) [3] The general logistic equation with small-population growth rate k0 and equi-librium population p is y = k0(1 − (y/p))y, The top menu choice is y˙ = (1 − y)y − a, which is the case k0 = 1 and p = 1 together with a hunt rate of a. The only added assumption is k0 = 1. (b) [3] 0 = (1 − y)y − a is the same as y2 − y + a = 0, which by the quadratic formula has solutions y = 21 ± 41 − a. Thus when a > 41 there are no equilibria; when a = 41 there is one, namely y0 = 12 , and it is semi-stable; and when a < 14 there are two, the top one stable and the bottom one unstable. 3 3 1 ± 1 1 3(c) [3] 187.5 oryx is 16 kilo-oryx, and a = 16 leads to critical points 2 4 or 4 and 4 . So the stable equilibrium population is 750 animals, and the critical population below which it will crash is 250. (d) [5] (e) [2] y2 − y − a = 0. 9. (a) [3] y0 = 3/4, from (b) above: y = u+3 , so 1−y = 1 −u and ˙u = (1 −u)(u−3 )− 3 = 4 4 4 4 16 −1 u − u2 . No explicit time dependence, so autonomous; and if u = 0 then ˙u = 0. 2 (b) [3] The linearized equation is ˙u = −12 u. The general solution to this is u = ce−t/2 . (c) [3] Thus y is well approximated by 43 + ce−t/2: the population decays, or relaxes, exponentially (with decay rate 21 ) to the equilibrium value. (d) [3] Both p(t) and q(t) must be constants.� � � � � � � � 11. (a) [4] p(s) = 12 s2 + 32 s + 58 = 12 (s2 + 3s + 54 ). One way to find the roots is by completing the square: s2 + 3s + 54 = (s + 32 )2 − 1, which clearly has roots −32 ± 1, or −12 and −52 . This is what is shown on the applet. (b) [4] x = c1e−t/2 + c2e−5t/2 . So x˙ = −12 c1e−t/2 − 52 c2e−5t/2, and x0 = c1 + c2, x˙0 = −12 c1 − 52 c2. Thus x0 +2 ˙x0 = −4c2 so c2 = −14 (x0 +2 ˙x0). Then c1 = x0 −c2 = 14 (5x0 +2 ˙x0). (c) [3] x is purely exponential when either c1 = 0—so 5x0 +2 ˙x0 = 0—or when c2 = 0—so x0 + 2 ˙x0 = 0. (d)[4] Try to solve for t in 0 = x(t) = c1e−t/2 + c2e−5t/2 . This leads to c2/c1 = −e2t . This admits a solution for some t exactly when c1 and c2 are of opposite sign. To get positive t, you need c2/c1 < −1: so either −c2 > c1 > 0 or −c2 < c1 < 0. In terms of x0, x˙0, this says either x0 + 2 ˙x0 > 5x0 + 2 ˙x0 > 0, or x0 + 2 ˙x0 < 5x0 + 2 ˙x0 < 0, i.e. either x0 < 0 and x˙0 > 52 (−x0), or x0 > 0 and ˙x0 < −25 x0. This is borne out by the applet. 12. (a) [6] p(s) = 12 (s2 + 2bs + 54 ) = 12 ((s + b)2 + (54 − b2)) has a double root when 5 = b2 or b = 5 . (We don’t allow b < 0.) Then the root is −b, so the general solution 4 2 is (a + ct)e−bt . (b) [6] When b = 1 , p(s) = 1 (s2 + 1 s + 5 ) = 1 ((s + 1 )2 + 19 ) has roots −1 ± 19 i �4 2 2 4 2 4 � 16 � 4 �� � � 4 −0.25 ± (1.0897)i. The general solution is thus e−t/4 a cos 419 t + b sin 419 t = Ae−t/4 cos 419 t − � . (Either form suffices.) (c) [5] My measurements are: 0.00, 2.93, 5.76, 8.69, 11.52. The successive differences are 2.93, 2.83, 2.93, 2.83—pretty close to constant. This is half the period of the sinusoid involved in the solution, which has circular frequency � = 419 and hence half-period � 4� = � � 2.8829231. Not bad agreement! The oscillations are constant over time� 19 (though the amplitude decreases). Successive differences of zeros of other solutions should be the same.MIT OpenCourseWare http://ocw.mit.edu 18.03 Differential Equations���� Spring 2010 For information about citing these materials or our Terms of Use, visit:
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