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

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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. μ ( 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 ) = 3 cos 2 t 4 t + 3 2 sin 2 t + c t As t → ∞ the first and second term go to zero and the solutions approach the function y s ( t ) = 3 sin 2 t/ 2. 5. The equation is in standard form and the coefficient p ( t ) = - 2 and the integrating factor is the following. μ ( 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. μ ( 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. μ ( t ) = e R 2 t dt = 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
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The solutions will approach zero as t → ∞ . 14. The equation is in standard form and the coefficient p ( t ) = 2 and the integrating factor is the following.
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