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283_pdfsam_math 54 differential equation solutions odd

283_pdfsam_math 54 differential equation solutions odd -...

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Exercises 5.2 Thus, a particular solution to the differential equation given in (5.23) is x p ( t ) = 90 . 4 and the general solution to this equation will be x ( t ) = c 1 e r 1 t + c 2 e r 2 t + 90 . 4 , where r 1 = ( 5 + 5) / 8 and r 2 = ( 5 5) / 8. To determine the maximum temperature of the attic, we will assume that zones A and B have sufficiently cool initial temperatures. (So that, for example, c 1 and c 2 are negative.) Since r 1 and r 2 are negative, as t goes to infinity, c 1 e r 1 t and c 2 e r 2 t each go to zero. Therefore, the maximum temperature that can be attained in the attic will be lim t →∞ x ( t ) = 90 . 4 F . 37. In this problem, we combine the idea exploded in interconnected tanks problems, rate of change = rate in rate out , (5.24) with the Newton’s law of cooling
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