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Math EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

The temperature of the classroom at noon and 5:00 P.M. The temperature of the classroom is 71 .8 ° F _ at noon. The temperature of the classroom is 26 .9 ° F _ at 5 P.M. Given: The classroom temperature T ( in units 100 ° F ) varies with time t ( in hours ) as d T d t = { 1 − T , if heating unit is ON . − T , if heating unit is OFF . Let T = 0 at 9:00 A.M., the heating unit is ON from 9-10 A.M., OFF from 10-11 A.M., ON again from 11 A.M.- noon, and so on for the rest of the day. Calculation: Let t = t 0 be the time when the heat is turned ON and T 0 be the temperature of the classroom at that time. That is, T ( t 0 ) = T 0 . Given that, the heating unit is turned ON at 9:00 A.M, 11:00 A.M, 1:00 P.M, and so on. Suppose the heating unit is turned ON, the temperature T ( in units 100 ° F ) varies with time t ( in hours ) as d T d t = 1 − T , T ( t 0 ) = T 0 d T d t + T = 1 , T ( t 0 ) = T 0 ... ( 1 ) Here, P ( t ) = 1 and Q ( t ) = 1 . Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 1 ) d t ] = e t Multiply the equation (1) by μ ( t ) and obtain d d t ( μ ( t ) T ( t ) ) = μ ( t ) Q ( t ) . e t ( d T d t + T ) = e t ( 1 ) e t d T d t + e t T = e t d d t ( e t T ) = e t To obtain the general solution T ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( e t T ) = e t and solve for T ( t ) by dividing μ ( t ) . e t T = ∫ e t d t e t T = e t + c T ( t ) = e t + c e t T ( t ) = 1 + c e − t Apply the initial condition T ( t 0 ) = T 0 in to T ( t ) = 1 + c e − t find the constant c . T ( t 0 ) = 1 + c e − t 0 T ( t 0 ) − 1 = c e − t 0 c = T ( t 0 ) − 1 e − t 0 c = ( T ( t 0 ) − 1 ) e t 0 Substitute c = ( T ( t 0 ) − 1 ) e t 0 in the solution T ( t ) = 1 + c e − t and simplify further. T ( t ) = 1 + [ T ( t 0 ) − 1 ] e t 0 e − t Thus, the temperature if the heating unit is turned ON can be written as T ( t ) = 1 + [ T ( t 0 ) − 1 ] e t 0 e − t ... ( 1 ) Let t = t 0 be the time when the heating unit is turned OFF and T 0 be the temperature of the classroom at that time. That is, T ( t 0 ) = T 0 . Given that, the heat is turned OFF at 10:00 A.M, Noon, 2:00 P.M, and so on. Suppose the heat is turned OFF, the temperature T ( in units 100 ° F ) varies with time t ( in hours ) as d T d t = − T , T ( t 0 ) = T 0 d T d t + T = 0 , T ( t 0 ) = T 0 ... ( 3 ) Here, P ( t ) = 1 and Q ( t ) = 0 . Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 1 ) d t ] = e t Multiply the equation (1) by μ ( t ) and obtain d d t ( μ ( t ) T ( t ) ) = μ ( t ) Q ( t ) . e t ( d T d t + T ) = e t ( 0 ) e t d T d t + e t T = 0 d d t ( e t T ) = 0 To obtain the general solution T ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( e t T ) = 0 and solve for T ( t ) by dividing μ ( t ) . e t T = ∫ ( 0 ) d t e t T = c T ( t ) = c e − t Apply the initial condition T ( t 0 ) = T 0 in to T ( t ) = c e − t find the constant c . T ( t 0 ) = c e − t 0 c = T ( t 0 ) e − t 0 Substitute c = T ( t 0 ) e − t 0 in the solution T ( t ) = c e − t and simplify further. T ( t ) = ( T ( t 0 ) e − t 0 ) e − t T ( t ) = T ( t 0 ) e t 0 e − t Thus, the temperature if the heat is turned OFF can be written as T ( t ) = T ( t 0 ) e t 0 e − t ... ( 4 ) Let the time t = 0 and the temperature T 0 = 0 at 9 A.M and the heat is turned ON. Thus by equation (2), the temperature becomes T ( t ) = 1 + [ T ( 0 ) − 1 ] e 0 e − t T ( t ) = 1 + ( 0 − 1 ) e − t T ( t ) = 1 − e − t The heating unit is turned OFF at 10 A.M. Take this time t = 1 and the corresponding temperature becomes T ( 1 ) = 1 − e − 1 Thus by equation (4), the temperature varies with time t from 10 A.M to 11 A.M is T ( t ) = T ( 1 ) e 1 e − t = ( 1 − e − 1 ) e 1 e − t = ( e − 1 ) e − t = e − t + 1 − e − t = − e − t + e − ( t − 1 ) The heating unit is turned ON at 11 A.M. Take this time t = 2 and the corresponding temperature becomes T ( 2 ) = − e − 2 + e − ( 2 − 1 ) = − e − 2 + e − 1 = e − 1 − e − 2 Thus by equation (2), the temperature varies with time t from 11 A.M to noon is T ( t ) = 1 + [ T ( 2 ) − 1 ] e ( 2 ) − t = 1 + ( e − 1 − e − 2 − 1 ) e 2 − t = 1 + e 1 − t − e − t − e 2 − t = 1 − e 2 − t + e 1 − t − e − t Proceeding like this way, the temperature corresponds to the time say t 1 when the heating unit is ON, we have T ( t 1 ) = ∑ n = 1 t 1 ( − 1 ) n + 1 e − n where t 1 = 2 , 4 , 6 , … ... ( 5 ) Also, the temperature corresponds to the time say t 2 when the heating unit is OFF, we have T ( t 2 ) = 1 + ∑ n = 1 t 2 ( − 1 ) n e − n where t 2 = 1 , 3 , 5 , … ... ( 6 ) At noon, the heating unit is OFF and the time is t 2 = 3 . Thus the temperature of the classroom at noon becomes T ( 3 ) = 1 + ∑ n = 1 3 ( − 1 ) n e − n = 1 + ( − 1 ) 1 e − 1 + ( − 1 ) 2 e − 2 + ( − 1 ) 3 e − 3 = 1 − e − 1 + e − 2 − e − 3 ≈ 0.718 Since the temperature T measured in 100 ° F , the temperature becomes T ( 3 ) = 0.718 × 100 ° F = 71 .8 ° F Therefore, the temperature of the classroom is 71 .8 ° F _ at noon. At 5 P.M., the heating unit is ON and the time is t 2 = 8 . Thus the temperature of the classroom at 5 P.M. becomes T ( 8 ) = ∑ n = 1 8 ( − 1 ) n + 1 e − n = e − 1 − e − 2 + e − 3 − e − 4 + e − 5 − e − 6 + e − 7 − e − 8 ≈ 0.269 Since the temperature T measured in 100 ° F , the temperature becomes T ( 8 ) = 0.269 × 100 ° F = 26 .9 ° F Therefore, the temperature of the classroom is 26 .9 ° F _ at 5 P.M.

The temperature of the classroom at noon and 5:00 P.M. The temperature of the classroom is 71 .8 ° F _ at noon. The temperature of the classroom is 26 .9 ° F _ at 5 P.M. Given: The classroom temperature T ( in units 100 ° F ) varies with time t ( in hours ) as d T d t = { 1 − T , if heating unit is ON . − T , if heating unit is OFF . Let T = 0 at 9:00 A.M., the heating unit is ON from 9-10 A.M., OFF from 10-11 A.M., ON again from 11 A.M.- noon, and so on for the rest of the day. Calculation: Let t = t 0 be the time when the heat is turned ON and T 0 be the temperature of the classroom at that time. That is, T ( t 0 ) = T 0 . Given that, the heating unit is turned ON at 9:00 A.M, 11:00 A.M, 1:00 P.M, and so on. Suppose the heating unit is turned ON, the temperature T ( in units 100 ° F ) varies with time t ( in hours ) as d T d t = 1 − T , T ( t 0 ) = T 0 d T d t + T = 1 , T ( t 0 ) = T 0 ... ( 1 ) Here, P ( t ) = 1 and Q ( t ) = 1 . Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 1 ) d t ] = e t Multiply the equation (1) by μ ( t ) and obtain d d t ( μ ( t ) T ( t ) ) = μ ( t ) Q ( t ) . e t ( d T d t + T ) = e t ( 1 ) e t d T d t + e t T = e t d d t ( e t T ) = e t To obtain the general solution T ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( e t T ) = e t and solve for T ( t ) by dividing μ ( t ) . e t T = ∫ e t d t e t T = e t + c T ( t ) = e t + c e t T ( t ) = 1 + c e − t Apply the initial condition T ( t 0 ) = T 0 in to T ( t ) = 1 + c e − t find the constant c . T ( t 0 ) = 1 + c e − t 0 T ( t 0 ) − 1 = c e − t 0 c = T ( t 0 ) − 1 e − t 0 c = ( T ( t 0 ) − 1 ) e t 0 Substitute c = ( T ( t 0 ) − 1 ) e t 0 in the solution T ( t ) = 1 + c e − t and simplify further. T ( t ) = 1 + [ T ( t 0 ) − 1 ] e t 0 e − t Thus, the temperature if the heating unit is turned ON can be written as T ( t ) = 1 + [ T ( t 0 ) − 1 ] e t 0 e − t ... ( 1 ) Let t = t 0 be the time when the heating unit is turned OFF and T 0 be the temperature of the classroom at that time. That is, T ( t 0 ) = T 0 . Given that, the heat is turned OFF at 10:00 A.M, Noon, 2:00 P.M, and so on. Suppose the heat is turned OFF, the temperature T ( in units 100 ° F ) varies with time t ( in hours ) as d T d t = − T , T ( t 0 ) = T 0 d T d t + T = 0 , T ( t 0 ) = T 0 ... ( 3 ) Here, P ( t ) = 1 and Q ( t ) = 0 . Calculate the integrating factor μ ( t ) by the formula μ ( t ) = exp [ ∫ P ( t ) d t ] . μ ( t ) = exp [ ∫ ( 1 ) d t ] = e t Multiply the equation (1) by μ ( t ) and obtain d d t ( μ ( t ) T ( t ) ) = μ ( t ) Q ( t ) . e t ( d T d t + T ) = e t ( 0 ) e t d T d t + e t T = 0 d d t ( e t T ) = 0 To obtain the general solution T ( t ) = 1 μ ( t ) [ ∫ μ ( t ) Q ( t ) d t + C ] , integrate both sides of the equation d d t ( e t T ) = 0 and solve for T ( t ) by dividing μ ( t ) . e t T = ∫ ( 0 ) d t e t T = c T ( t ) = c e − t Apply the initial condition T ( t 0 ) = T 0 in to T ( t ) = c e − t find the constant c . T ( t 0 ) = c e − t 0 c = T ( t 0 ) e − t 0 Substitute c = T ( t 0 ) e − t 0 in the solution T ( t ) = c e − t and simplify further. T ( t ) = ( T ( t 0 ) e − t 0 ) e − t T ( t ) = T ( t 0 ) e t 0 e − t Thus, the temperature if the heat is turned OFF can be written as T ( t ) = T ( t 0 ) e t 0 e − t ... ( 4 ) Let the time t = 0 and the temperature T 0 = 0 at 9 A.M and the heat is turned ON. Thus by equation (2), the temperature becomes T ( t ) = 1 + [ T ( 0 ) − 1 ] e 0 e − t T ( t ) = 1 + ( 0 − 1 ) e − t T ( t ) = 1 − e − t The heating unit is turned OFF at 10 A.M. Take this time t = 1 and the corresponding temperature becomes T ( 1 ) = 1 − e − 1 Thus by equation (4), the temperature varies with time t from 10 A.M to 11 A.M is T ( t ) = T ( 1 ) e 1 e − t = ( 1 − e − 1 ) e 1 e − t = ( e − 1 ) e − t = e − t + 1 − e − t = − e − t + e − ( t − 1 ) The heating unit is turned ON at 11 A.M. Take this time t = 2 and the corresponding temperature becomes T ( 2 ) = − e − 2 + e − ( 2 − 1 ) = − e − 2 + e − 1 = e − 1 − e − 2 Thus by equation (2), the temperature varies with time t from 11 A.M to noon is T ( t ) = 1 + [ T ( 2 ) − 1 ] e ( 2 ) − t = 1 + ( e − 1 − e − 2 − 1 ) e 2 − t = 1 + e 1 − t − e − t − e 2 − t = 1 − e 2 − t + e 1 − t − e − t Proceeding like this way, the temperature corresponds to the time say t 1 when the heating unit is ON, we have T ( t 1 ) = ∑ n = 1 t 1 ( − 1 ) n + 1 e − n where t 1 = 2 , 4 , 6 , … ... ( 5 ) Also, the temperature corresponds to the time say t 2 when the heating unit is OFF, we have T ( t 2 ) = 1 + ∑ n = 1 t 2 ( − 1 ) n e − n where t 2 = 1 , 3 , 5 , … ... ( 6 ) At noon, the heating unit is OFF and the time is t 2 = 3 . Thus the temperature of the classroom at noon becomes T ( 3 ) = 1 + ∑ n = 1 3 ( − 1 ) n e − n = 1 + ( − 1 ) 1 e − 1 + ( − 1 ) 2 e − 2 + ( − 1 ) 3 e − 3 = 1 − e − 1 + e − 2 − e − 3 ≈ 0.718 Since the temperature T measured in 100 ° F , the temperature becomes T ( 3 ) = 0.718 × 100 ° F = 71 .8 ° F Therefore, the temperature of the classroom is 71 .8 ° F _ at noon. At 5 P.M., the heating unit is ON and the time is t 2 = 8 . Thus the temperature of the classroom at 5 P.M. becomes T ( 8 ) = ∑ n = 1 8 ( − 1 ) n + 1 e − n = e − 1 − e − 2 + e − 3 − e − 4 + e − 5 − e − 6 + e − 7 − e − 8 ≈ 0.269 Since the temperature T measured in 100 ° F , the temperature becomes T ( 8 ) = 0.269 × 100 ° F = 26 .9 ° F Therefore, the temperature of the classroom is 26 .9 ° F _ at 5 P.M.

Solution Summary: The author calculates the temperature of the classroom at noon and 5:00 P.M.

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

9th Edition

ISBN: 9780135963777

Author: Nagle

Publisher: PEARSON CO

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EP FUND.OF DIFF.EQUATIONS-MYLAB (18 WK)

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In Example 2 assume that the...Ch. 2.3 - Prob. 24E Ch. 2.3 - (a) Using definite integration, show that the...Ch. 2.3 - Prob. 26E Ch. 2.3 - Constant Multiples of Solutions. (a) Show that y =...Ch. 2.3 - Prob. 29E Ch. 2.3 - Bernoulli Equations. 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(x2 + y2)...Ch. 2 - Prob. 10RP Ch. 2 - Prob. 11RP Ch. 2 - Prob. 12RP Ch. 2 - Prob. 13RP Ch. 2 - Prob. 14RP Ch. 2 - Prob. 15RP Ch. 2 - Prob. 16RP Ch. 2 - Prob. 17RP Ch. 2 - Prob. 18RP Ch. 2 - Prob. 19RP Ch. 2 - Prob. 20RP Ch. 2 - Prob. 21RP Ch. 2 - Prob. 22RP Ch. 2 - Prob. 23RP Ch. 2 - In Problems 130, solve the equation. 24. (y/x +...Ch. 2 - Prob. 25RP Ch. 2 - Prob. 26RP Ch. 2 - Prob. 27RP Ch. 2 - Prob. 28RP Ch. 2 - Prob. 29RP Ch. 2 - Prob. 30RP Ch. 2 - Prob. 31RP Ch. 2 - Prob. 32RP Ch. 2 - Prob. 33RP Ch. 2 - Prob. 34RP Ch. 2 - Prob. 35RP Ch. 2 - Prob. 36RP Ch. 2 - Prob. 37RP Ch. 2 - Prob. 38RP Ch. 2 - Prob. 39RP Ch. 2 - Prob. 40RP Ch. 2 - Prob. 41RP

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