120_pdfsam_math 54 differential equation solutions odd

120_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 Integrating, we obtain Bu( t )=2 Z ( e t e 2 t ) dt = 2 e t + e 2 t + C, and the initial condition Bu(0) = 0 gives C = 1. Therefore, Bu( t )= 2 e t + e 2 t +1 . EXERCISES 3.3: Heating and Cooling of Buildings, page 107 1. Let T ( t ) denote the temperature of coFee at time t (in minutes). According to the Newton’s Law (1) on page 102 of the text, dT dt = K [21 T ( t )] , wherewehavetaken H ( t ) U ( t ) 0, M ( t ) 21 C, with the initial condition T (0) = 95 C. Solving this initial value problem yields dT 21 T = Kdt ⇒− ln | T 21 | = Kt + C 1 T ( t )=21+ Ce ; 95 = T (0) = 21 + K (0) C =74 T ( t )=21+74 e . To ±nd K , we use the fact that after 5 min the temperature of coFee was 80 C. Thus 80 = T (5) = 21 + 74 e K (5) K = ln(74 / 59) 5 , and so
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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 Berkeley.

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