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41_pdfsam_math 54 differential equation solutions odd

41_pdfsam_math 54 differential equation solutions odd -...

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Exercises 2.2 Therefore, dA dt = 3 A 40 = 120 A 40 . Separating this differential equation and integrating yield 40 120 A dA = dt 40 ln | 120 A | = t + C ln | 120 A | = t 40 + C, where C 40 is replaced by C 120 A = Ce t/ 40 , where C can now be positive or negative A = 120 Ce t/ 40 . There are 2 kg of salt in the tank initially, thus A (0) = 2. Using this initial condition, we find 2 = 120 C C = 118 . Substituting this value of C into the solution, we have A = 120 118 e t/ 40 . Thus A (10) = 120 118 e 10 / 40 28 . 1 kg . Note: There is a detailed discussion of mixture problems in Section 3.2. 35. In Problem 34 we saw that the differential equation
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