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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 semi-stable—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...
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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