281_pdfsam_math 54 differential equation solutions odd

281_pdfsam_math 54 differential equation solutions odd -...

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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 Newtons 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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