TEST2solutions

TEST2solutions - Math 527, Test #2(SAMPLE Solutions)...

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Unformatted text preview: Math 527, Test #2(SAMPLE Solutions) University of New Hampshire Fall 2007 NAME: Section: PART 1 - Closed Notes. Please complete any 3 of the 4 problems provided. Cross out the problem you do not want graded in the table below. In cases of ambiguity, we will grade the last three problems . This portion of the exam is closed calculator, closed notes and closed neighbor. No collaboration is permitted. If you have any questions, please raise your hand. Question Points Possible 1 20 2 20 3 20 4 20 1 1. Solve each of the following homogeneous equations for their general solutions. (a) y 00- 2 y- 8 y = 0 Assuming an exponential solution, y = ke rt the characteristic polynomial be- comes: r 2- 2 r- 8 = 0 Solving this quadratic we find distinct, real roots: r = 4 ,- 2 . This yields the general solution: y = C 1 e 4 t + C 2 e- 2 t (b) y 00 + 2 y + 10 y = 0 Assuming an exponential solution, y = ke rt the characteristic polynomial be- comes: r 2 + 2 r + 10 = 0 Solving this quadratic using the quadratic formula we find complex roots: r =- 2 ± √ 4- 40 2 r =- 1 ± 3 i This yields the general solution: y = C 1 e- t cos(3 t ) + C 2 e- t sin(3 t ) (c) y 00 + 6 y + 9 y = 0 Assuming an exponential solution, y = ke rt the characteristic polynomial be- comes: r 2 + 6 r + 9 = 0 Solving this quadratic using the quadratic formula we find a repeated root: r =- 3 . This yields the general solution: y = C 1 e- 3 t + C 2 te- 3 t 2 (d) t 2 y 00 + 4 ty + 2 y = 0 Here we have an Euler equation so we make the guess y = at r . Using this guess we obtain the characteristic polynomial: t 2 r ( r- 1) t r- 2 + 4 trt r- 1 + 2 t r = 0 r ( r- 1) + 4 r + 2 = 0 r 2 + 3 r + 2 = 0 Solving for the roots we find r =- 2 ,- 1 . We obtain the general solution: y = C 1 t- 2 + C 2 t- 1 3 2. Solve the following homogeneous initial value problems: (a) y 00 + y = 0, y (0) = 1, y (0) = 0 The characteristic polynomial here is: r 2 + 1 = 0 We obtain pure imaginary roots, r = ± i and we use trig functions for our general solution....
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This note was uploaded on 04/08/2008 for the course MATH 527 taught by Professor Boucher during the Fall '07 term at New Hampshire.

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TEST2solutions - Math 527, Test #2(SAMPLE Solutions)...

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