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Problem Set 3, Population Growth-Differential Equations-Assignment Solution, Exercises of Differential Equations

Institute of Mathematics and ApplicationsDifferential Equations

Differentiation Equations course is one of basic course of science study. Its part of Mathematics, Computer Science, Physics, Engineering. This is assignment solution to help students for exercising problem. It includes: Problem, Set, Poulation, Growth, General, Logistic, Equation, Stable, Equilibrium, Critical, Linearized, Applet, Completing, Square, Root

Typology: Exercises

2011/2012

Uploaded on 08/07/2012

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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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18.03 Problem Set 3: Part II Solutions

Part I points: 8. 8, 9. 12, 11. 9, 12. 7.

  1. (a) [3] The general logistic equation with small-population growth rate k 0 and equi librium population p is y = k 0 (1 − (y/p))y, The top menu choice is y˙ = (1 − y)y − a, which is the case k 0 = 1 and p = 1 together with a hunt rate of a. The only added assumption is k 0 = 1. (b) [3] 0 = (1 − y)y − a is the same as y^2 − 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 y 0 = 12 , and it is semi-stable; and when a < 14 there are two, the top one stable and the bottom one unstable. (c) [3] 187. 5 oryx is 3 3 1 ± 1 1 3 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] y^2 − y − a = 0.
  2. (a) [3] y 0 = 3/4, from (b) above: y = u+ 34 , so 1 −y = 14 −u and u˙ = ( 14 −u)(u− 34 )− 163 = −^12 u − u^2. No explicit time dependence, so autonomous; and if u = 0 then u˙ = 0. (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.

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� � � � � (^) � � �

  1. (a) [4] p(s) = 12 s^2 + 32 s + 58 = 12 (s^2 + 3s + 54 ). One way to find the roots is by completing the square: s^2 + 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 = c 1 e−t/^2 + c 2 e−^5 t/^2. So x˙ = − 12 c 1 e−t/^2 − 52 c 2 e−^5 t/^2 , and x 0 = c 1 + c 2 , x˙ 0 = −^12 c 1 − 52 c 2. Thus x 0 +2 x˙ 0 = − 4 c 2 so c 2 = − 14 (x 0 +2 x˙ 0 ). Then c 1 = x 0 −c 2 = 14 (5x 0 +2 x˙ 0 ). (c) [3] x is purely exponential when either c 1 = 0—so 5 x 0 + 2 ˙x 0 = 0—or when c 2 = 0—so x 0 + 2 ˙x 0 = 0. (d)[4] Try to solve for t in 0 = x(t) = c 1 e−t/^2 + c 2 e−^5 t/^2. This leads to c 2 /c 1 = −e^2 t^. This admits a solution for some t exactly when c 1 and c 2 are of opposite sign. To get positive t, you need c 2 /c 1 < −1: so either −c 2 > c 1 > 0 or −c 2 < c 1 < 0. In terms of x 0 , x˙ 0 , this says either x 0 + 2 ˙x 0 > 5 x 0 + 2 ˙x 0 > 0, or x 0 + 2 ˙x 0 < 5 x 0 + 2 ˙x 0 < 0, i.e. either x 0 < 0 and x˙ 0 > 52 (−x 0 ), or x 0 > 0 and x˙ 0 < − 25 x 0. This is borne out by the applet.
  2. (a) [6] p(s) = 12 (s^2 + 2bs + 54 ) = 12 ((s + b)^2 + (^54 − b^2 )) has a double root when (^5) = b (^2) 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 = 14 , p(s) = 12 (s^2 + 12 s + 54 ) = 12 ((s + (^14) �^ )^2 + (^1916) �^ ) has� roots − (^) �^14 ± (^) �� 419 i � − 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
  3. 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.

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