Unformatted text preview: Exercises 5.2 Next, we use the initial condition, x (0) = 0, y (0) = 20, to find values of c 1 and c 2 . 20 − (2 + √ 5) c 1 + ( − 2 + √ 5) c 2 = 0 , 20 + c 1 + c 2 = 20 ⇒ c 1 = 10 / √ 5 , c 2 = − 10 / √ 5 . With these values, the solution given in (5.21), (5.22) becomes x ( t ) = 20 − 20 + 10 √ 5 √ 5 e ( − 3 − √ 5) t/ 100 + 20 − 10 √ 5 √ 5 e ( − 3+ √ 5) t/ 100 , y ( t ) = 20 + 10 √ 5 e ( − 3 − √ 5) t/ 100 − 10 √ 5 e ( − 3+ √ 5) t/ 100 . 35. Let x ( t ) and y ( t ) denote the temperatures at time t in zones A and B , respectively. Therefore, the rate of change of temperature in zone A will be x ( t ) and in zone B will be y ( t ). We can apply Newton’s law of cooling to help us express these rates of change in an alternate manner. Thus, we observe that the rate of change of the temperature in zone A due to the outside temperature is k 1 [100 − x ( t )] and due to the temperature in zone B is k 2 [ y ( t ) − x ( t...
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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