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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 = 2
1 ± 4
1 − a. Thus when a > 4
1 there are no equilibria; when a = 4
1
there is one, namely y0 = 1
2 , and it is semi-stable; and when a < 1
4 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 = −1
2 u. The general solution to this is u = ce−t/2
.
(c) [3] Thus y is well approximated by 4
3 + ce−t/2: the population decays, or relaxes,
exponentially (with decay rate 2
1 ) to the equilibrium value.
(d) [3] Both p(t) and q(t) must be constants.
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