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Unformatted text preview: Exercises 5.2
Thus, a particular solution to the diﬀerential equation given in (5.23) is xp (t) = 90.4 and the general solution to this equation will be x(t) = c1 er1 t + c2 er2 t + 90.4 , where r1 = (−5 + √ 5)/8 and r2 = (−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, c1 and c2 are negative.) Since r1 and r2 are negative, as t goes to inﬁnity, c1 er1 t and c2 er2 t each go to zero. Therefore, the maximum temperature that can be attained in the attic will be
t→∞ lim 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, with the Newton’s law of cooling dT = K (T − M ). dt Let x(t) and y (t) denote temperatures in rooms A and B, respectively. Room A. It gets temperature only from the heater with a rate rate in = 80, 000 Btu/h · 1/4◦ = 20◦ /h. 1000 Btu (5.25) (5.24) Temperature goes out of the room A into the room B and outside with diﬀerent coeﬃcients of proportionality in (5.25): K1 = 1/2 and K2 = 1/4, respectively. Therefore, rate out = rate into B + rate outside 1 3 1 1 (x − y ) + (x − 0) = x − y. = 2 4 4 2 Thus, (5.24) implies that x = 20 − 3 1 x− y 4 2 = 20 − 3 1 x + y. 4 2 279 ...
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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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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