This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 18.03 Problem Set 3: Solutions 8. (F 22 Feb) (a) ˙ y = . 25 + ay y 2 = ( a y ) y . 25 represents this situation. When y is small, y 2 is very small, so when the hunting is turned off the equation is close to ˙ y = ay , representing a natural growth rate of a . (b) By the quadratic formula, ˙ y = 0 has a solution provided that a ≥ 1. When a = 1 there is a double root, y = 1 / 2. This is a dangerous strategy because the critical point y = 1 / 2 is semistable—if the population falls below it by chance then it will collapse to zero. (c) We want a such that y = 1 . 5 is a critical point. Thus 0 = ˙ y = . 25 + a (1 . 5) (1 . 5) 2 or a = 5 / 3. (d) (e) . 25 + ay y 2 = 0. 9. (M 25 Feb) (a) Separate: dy (1 ( y/p )) y = k dt . To integrate the left hand side we need to use partial fractions: 1 (1 ( y/p )) y = 1 y + 1 p y . Integrate: ln  y   ln  p y  = k t + c , or ln  y p y  = k t + c . Exponentiate and eliminate the absolute values: y p y = Ce k t . Solve for y : Multiply through by...
View
Full
Document
This note was uploaded on 02/06/2009 for the course 18 18.03 taught by Professor Unknown during the Spring '09 term at MIT.
 Spring '09
 unknown

Click to edit the document details