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Unformatted text preview: Math 527, Test #1SAMPLE(Solutions) University of New Hampshire Fall 2007 NAME: Please complete any 4 of the five problems provided. Cross out the problem you do not want graded in the table below. In cases of ambiguity, we will grade the first four prob lems. This exam is closed calculator, closed notes and closed neighbor. No collaboration is permitted. If you have any questions, please raise your hand. (Note: Problem 1 is spread over multiple pages) Question Points Possible 1 25 2 25 3 25 4 25 5 25 1 1. (a) Match each of the following differential equations to the appropriate slope field. i. y = y ii. y = y 2 1 iii. y = cos( t ) iv. y = y 1 ——–ii——— ———iv——— ——–iii——— ——–i——— 2 (b) Consider the following autonomous differential equation: dy dt = y (100 y ) 2 i. What are the equilibrium solutions for this autonomous equation? We obtain equilibrium solutions by setting dy dt = 0 and solving the resulting equation. Here the solutions are given by y = 0 , y = 100 ii. Draw a stability diagram for this equation and indicate the stability of each equilibrium solution. iii. Assume that this equation governs the dynamics of a particular spider pop ulation. Recently the population size was estimated around 90 spiders. What does this model predict will happen to the size of this population in the absence of any other factors? The model predicts that the population will rise asyptotically to 100 spiders and stay at that level (barring any unforseen changes) iv. If say, 50 spiders of the same species were introduced into this population what would happen (according to the model)? With the introduction of 50 extra spiders the population would rise above the equilibrium level of 100 spiders, so we see from the stability diagram that the rate of change of the population is positive in that region, which will lead to a population explosion. 3 2. (a) Consider the following initial value problem: dy dt 2 y = t, y ( π ) = 1 i. Solve this problem and find the particular solution. The integrating factor is given by e 2 t . Multiplying by this factor we obtain: d dt e 2 t y = te 2 t Now we integrate (using integration by parts on the right hand side), solve and apply the initial condition: Z d dt e 2 t y dt...
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This homework help was uploaded on 04/08/2008 for the course MATH 527 taught by Professor Boucher during the Fall '07 term at New Hampshire.
 Fall '07
 Boucher
 Math, Differential Equations, Equations

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