121_pdfsam_math 54 differential equation solutions odd

121_pdfsam_math 54 - Exercises 3.3 where M(t = 32 is the temperature of ice This equation is linear and is rewritten in the standard form as dT

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Exercises 3.3 where M ( t ) = 32 is the temperature of ice. This equation is linear and is rewritten in the standard form as dT dt + KT ( t )=32 K. We Fnd that the integrating factor is e Kt . Multiplying both sides by e Kt and integrating gives e Kt dT dt + e Kt KT ( t )=32 Ke Kt e Kt T ( t )= Z 32 Ke Kt dt e Kt T ( t )=32 e Kt + C T ( t )=32+ Ce Kt . By setting t = 0 and using the initial temperature 70 ±, we Fnd the constant C . 70 = 32 + C C =38 . Knowing that it takes 15 minutes for the wine to chill to 60 ±, we can Fnd the constant, K : 60 = 32 + 38 e K (15) . Solving for K yields K = 1 15 ln ± 60 32 38 ² 0 . 02035 . Therefore, T ( t )=32+38 e 0 . 02035 t . We can now determine how long it will take for the wine to reach 56
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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