# Hw3 - Homework#3 Section 2.2 3 dy dt y = 3 e t The integrating factor is given by μ t = e R 1 dt = e t Then e t y y = 3 ⇒ d dt ye t = 3 ⇒ ye t

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Unformatted text preview: Homework #3 Section 2.2 3. dy dt- y = 3 e t The integrating factor is given by μ ( t ) = e R (- 1) dt = e- t . Then, e- t ( y- y ) = 3 ⇒ d dt ( ye- t ) = 3 ⇒ ye- t = 3 t + c ⇒ y = 3 te t + ce t . 4. dy dt + y = sin t The integrating factor is given by μ ( t ) = e t . Then, e t ( y + y ) = e t sin t ⇒ d dt ( ye t ) = e t sin t ⇒ ye t = 1 2 e t (sin t- cos t ) + c ⇒ y = 1 2 (sin t- cos t ) + ce- t . 12. dy dt + 3 t y = sin t t 3 , ( t 6 = 0) The integrating factor is given by μ ( t ) = e R 3 t dt = e 3ln t = t 3 . Then, t 3 ( y + 3 t y ) = sin t ⇒ d dt ( t 3 y ) = sin t ⇒ t 3 y =- cos t + c ⇒ y = c t 3- cos t t 3 . 20. (1 + e t ) dy dt + e t y = 0, y (0) = 1 The differential equation can be rewritten as y + e t 1 + e t y = 0. Then, the integrating factor is given by μ ( t ) = e R e t 1+ e t dt = e ln(1+ e t ) = 1 + e t . Thus, d dt ( (1 + e t ) y ) = 0 ⇒ ( (1 + e t ) y ) = c ⇒ y = c 1 + e t ....
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## This note was uploaded on 03/23/2010 for the course APPM 2360 taught by Professor Williamheuett during the Spring '07 term at Colorado.

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Hw3 - Homework#3 Section 2.2 3 dy dt y = 3 e t The integrating factor is given by μ t = e R 1 dt = e t Then e t y y = 3 ⇒ d dt ye t = 3 ⇒ ye t

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