solutions1 - 1.1 15. j 16. c 17. g 18. b 19. h 20. e 2.1 4....

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Unformatted text preview: 1.1 15. j 16. c 17. g 18. b 19. h 20. e 2.1 4. The equation is in standard form and the coefficient p ( t ) = 1 /t and the integrating factor is the following. ( t ) = e R dt t = e ln | t | = t Using this integrating factor in the general formula on page 36 gives us the solution. y ( t ) = 3cos2 t 4 t + 3 2 sin2 t + c t As t the first and second term go to zero and the solutions approach the function y s ( t ) = 3sin2 t/ 2. 5. The equation is in standard form and the coefficient p ( t ) =- 2 and the integrating factor is the following. ( t ) = e- R 2 dt = e- 2 t Using this integrating factor in the general formula on page 36 gives us the solution. y ( t ) =- 3 e t + ce 2 t The solutions will grow exponentially as t . 6. After rewriting the equation in standard form, the coefficient p ( t ) = 2 /t and the integrating factor is the following. ( t ) = e R 2 dt t = e 2 ln | t | = t 2 Using this integrating factor in the general formula on page 36 gives us the solution. y ( t ) =- cos t t + sin t t 2 + c t 2 The solutions will approach zero as t . 7. The equation is in standard form and the coefficient p ( t ) = 2 t and the integrating factor is the following. ( t ) = e R 2 tdt = e t 2 Using this integrating factor in the general formula on page 36 gives us the solution. y ( t ) = t 2 e- t 2 + ce- t 2 The solutions will approach zero as t ....
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This note was uploaded on 12/29/2010 for the course MATH 265 taught by Professor Leahkeshet during the Winter '10 term at The University of British Columbia.

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solutions1 - 1.1 15. j 16. c 17. g 18. b 19. h 20. e 2.1 4....

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