Explanation Given information: Assume that the outside temperature varies as T ( t ) = 15 + 5 s i n ( π t 12 ) Where t = 0 is 12 noon. A house is heated to 25 o C at t = 0 and after that, its temperature y ( t ) varies according to Newton’s Law of Cooling: d y d t = − 0.1 ( y ( t ) − T ( t ) ) Formula used: If a , b and r be constants and if d y d t = − k ( y − a − b sin r t ) then its general solution is y = C e − k t + a + b k ( k sin r t − r cos r t k 2 + r 2 ) . Calculation: From the given data, we have T ( t ) = 15 + 5 s i n ( π t 12 ) (1) d y d t = − 0.1 ( y ( t ) − T ( t ) ) (2) Using (1) in (2), we get d y d t = − 0.1 ( y ( t ) − ( 15 + 5 s i n ( π t 12 ) ) ) d y d t = − 0.1 ( y ( t ) − 15 − 5 s i n ( π t 12 ) ) The above equation is of the form d y d t = − k ( y − a − b sin r t ) with k = 0.1 , a = 15 , b = 5 , r = π 12 ,then its general solution will be y ( t ) = C e − 0.1 t + 15 + 5 ( 0.1 ) ( 0.1 sin ( π 12 ) t − ( π 12 ) cos ( π 12 ) t ( 0.1 ) 2 + ( π 12 ) 2 ) y ( t ) = C e − 0

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Chapter 10.5, Problem 46E

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To solve:The differential equation dydt=−0.1(y(t)−T(t)) for y(t).

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Chapter 10.5, Problem 45E

Chapter 10.5, Problem 47E

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Chapter 10 Solutions

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The differential equation d y d t = − 0.1 ( y ( t ) − T ( t ) ) for y ( t ) .

Solution Summary: The author explains how to solve the differential equation dyt=-0.1(y(t)), based on Newton’s Law of Cooling.

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To solve:The differential equation dydt=−0.1(y(t)−T(t)) for y(t).

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Chapter 10 Solutions