22B 3.3-3.2

# 22B 3.3-3.2 - Math 22B Solutions Homework 4 Spring 2008...

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Unformatted text preview: Math 22B Solutions Homework 4 Spring 2008 Section 3.1 6. Solution Let y = e t , then 4 d 2 ( e t ) dt- 9 e t = 0 becomes 4 2- 9 = 0. So then 2 = 9 4 and = 3 2 . The general solution is y = c 1 e- 3 t 2 + c 2 e 3 t 2 . 16. Solution Let y = e t . The differential equation becomes 4 2- 1 = 0. Hence 2 = 1 4 and = 1 2 . The general solution becomes y = c 1 e- t 2 + c 2 e t 2 . Now, we have y (- 2) = c 1 e 1 + c 2 e- 1 = 1 and y (- 2) = c 1 (- 1 2 ) e 1 + c 2 ( 1 2 ) e- 1 =- 1. Thus, c 2 =- e 2 and c 1 = 3 2 e . So the general solution is y ( t ) = 3 2 e e- t 2- e 2 e t 2 . 21. Solution Let y = e t . The differential equation becomes 2- - 2 = ( - 2)( + 1) = 0. So = 2 or =- 1. The general solution is y = c 1 e 2 t + c 2 e- t . Now, y (0) = c 1 + c 2 = and y (0) = 2 c 1- c 2 = 2 gives us 3 c 1 = + 2. If y 0 as t then c 1 = 0 and =- 2. Hence, c 2 =- 2 and y =- 2 e- t ....
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## This note was uploaded on 05/13/2008 for the course MATH 22B taught by Professor Hunter during the Spring '08 term at UC Davis.

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22B 3.3-3.2 - Math 22B Solutions Homework 4 Spring 2008...

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