125_pdfsam_math 54 differential equation solutions odd

125_pdfsam_math 54 differential equation solutions odd - )]...

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Exercises 3.3 For the case when the time constant for the building is 1, we see that 1 /K = 1 which implies that K = 1. Therefore, the temperature will reach a maximum of K T =24+8 ± 1+ ² π 12 ³ 2 ´ 1 / 2 31 . 7 C . It will reach a minimum of T =24 8 ± 1+ ² π 12 ³ 2 ´ 1 / 2 16 . 3 C . For the case when the time constant of the building is 5, we have 1 K =5 K = 1 5 . Then, the temperature will reach a maximum of T =24+8 1+ µ 5 π 12 2 1 / 2 28 . 9 C , and a minimum of T =24 8 1+ µ 5 π 12 2 1 / 2 19 . 1 C . 11. As in Example 3, page 105 of the text, this problem involves a thermostat to regulate the temperature in the van. Hence, we have U ( t )= K U [ T D T ( t
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Unformatted text preview: )] , where T D is the desired temperature 16 C and K U is a proportionality constant. We will assume that H ( t ) = 0 and that the outside temperature M ( t ) is a constant 35 C. The time constant for the van is 1 /K = 2 hr, hence K = 0 . 5. Since the time constant for the van with its air conditioning system is 1 /K 1 = 1 / 3 hr, then K 1 = K + K U = 3. Therefore, K U = 3 . 5 = 2 . 5. The temperature in the van is governed by the equation dT dt = (0 . 5)(35 T ) + (2 . 5)(16 T ) = 57 . 5 3 T. 121...
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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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