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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 + π 12 2 1 / 2 31 . 7 C . It will reach a minimum of T = 24 8 1 + π 12 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 + 5 π 12 2 1 / 2 28 . 9 C , and a minimum of T = 24 8 1 + 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 )] , where T D
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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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