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126_pdfsam_math 54 differential equation solutions odd

126_pdfsam_math 54 differential equation solutions odd - H...

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Chapter 3 Solving this separable equation yields T ( t ) = 19 . 17 + Ce 3 t . When t = 0 we are given T (0) = 55. Using this information to solve for C gives C = 35 . 83. Hence, the van temperature is given by T ( t ) = 19 . 17 + 35 . 83 e 3 t . To find out when the temperature in the van will reach 27 C, we let T ( t ) = 27 and solve for t . Thus, we see that 27 = 19 . 17 + 35 . 83 e 3 t e 3 t = 7 . 83 35 . 83 0 . 2185 t ln(0 . 2185) 3 0 . 5070 (hr) or 30 . 4 min . 13. Since the time constant is 64, we have K = 1 / 64. The temperature in the tank increases at the rate of 2 F for every 1000 Btu. Furthermore, every hour of sunlight provides an input of 2000 Btu to the tank. Thus,
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Unformatted text preview: H ( t ) = 2 × 2 = 4 ◦ ± per hr . We are given that T (0) = 110, and that the temperature M ( t ) outside the tank is a constant 80 ◦ ±. Hence the temperature in the tank is governed by dT dt = 1 64 [80 − T ( t )] + 4 = − 1 64 T ( t ) + 5 . 25 , T (0) = 110 . Solving this separable equation gives T ( t ) = 336 + Ce − t/ 64 . To Fnd C , we use the initial condition to see that T (0) = 110 = 336 + C ⇒ C = − 226 . This yields the equation T ( t ) = 336 − 226 e − t/ 64 . 122...
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