TEST2solutions

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

This preview shows pages 1–5. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

## 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.

### Page1 / 12

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online