Exercises 5.2Thus, a particular solution to the differential equation given in (5.23) isxp(t) = 90.4 and thegeneral solution to this equation will bex(t) =c1er1t+c2er2t+ 90.4,wherer1= (−5 +√5)/8 andr2= (−5−√5)/8. To determine the maximum temperature ofthe attic, we will assume that zonesAandBhave suﬃciently cool initial temperatures. (Sothat, for example,c1andc2are negative.) Sincer1andr2are negative, astgoes to infinity,c1er1tandc2er2teach go to zero. Therefore, the maximum temperature that can be attainedin the attic will belimt→∞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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