22B 2.1-2.4

# 22B 2.1-2.4 - Math 22B Solutions Homework 2 Spring 2008...

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Unformatted text preview: Math 22B Solutions Homework 2 Spring 2008 Section 2.1 16. y + 2 t y = cos t t 2 , with y ( π ) = 0 and t = 0 Solution Let μ ( t ) = e R 2 t dt = e t 2 . If we multiply both sides of the given equation by μ ( t ), we get: t 2 ( y ) + 2 t = cos t ( t 2 y ) = cos t Then integrate both sides to get: t 2 y = sin t + c y = 1 t 2 sin t + c t 2 Setting initial values we get: y = 1 π 2 sin π + c π 2 = 0 y = 0 + c π 2 = 0 ⇒ c = 0 y = 1 t 2 sin t 22. (b) 2 y- y = e t 3 , with y (0) = a Solution μt = e- t 2 2 e- t 2 y- e- t 2 y = e t 3- t 2 y ( t ) =- 3 e t 3 + ce t 2 1 y (0) =- 3 + c = a ⇒ c = a + 3 y ( t ) =- 3 e t 3 + ( a + 3) e t 2 Differentiate to find critical points: y ( t ) =- e t 3 + ( a + 3) 2 e t 2 y (0) =- 1 + ( a + 3) 2 = ( a + 1) 2 , a =- 1 is a critical value. (c) Solution For g =- 1, y ( t ) =- 3 e t 2 + 2 e t 2 . This is dominated by e t 2 . 30. y- y = 1 + 3 sin t Solution μt = e- t ( e- t y ) = e- t + 3 e- t sin t e- t y =- e- t + 3 2 (- e- t cos t- e- t sin t ) + c y =- 1- 3 2 e- t (cos t + sin t ) + ce t If y ( t ) is bounded, c = 0, and y (0) =- 1- 3 2 + c = y Where c = y + 5 2 = 0 and...
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22B 2.1-2.4 - Math 22B Solutions Homework 2 Spring 2008...

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