41_pdfsam_math 54 differential equation solutions odd

41_pdfsam_math 54 differential equation solutions odd -...

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Unformatted text preview: Exercises 2.2 dA A 120 − A =3− = . dt 40 40 Separating this differential equation and integrating yield 40 dA = dt 120 − A ⇒ ⇒ ⇒ ⇒ −40 ln |120 − A| = t + C Therefore, ln |120 − A| = − C t + C, where − is replaced by C 40 40 120 − 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 − 118e−t/40 . Thus A(10) = 120 − 118e−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 dT /dt = k (M − T ) can be solved by separation of variables to yield T = Cekt + M. When the oven temperature is 120◦ we have M = 120. Also T (0) = 40. Thus 40 = C + 120 Because T (45) = 90, we have 90 = −80e45k + 120 ⇒ 3 = e45k 8 ⇒ 45k = ln 3 . 8 ⇒ C = −80. Thus k = ln(3/8)/45 ≈ −0.02180. This k is independent of M . Therefore, we have the general equation T (t) = Ce−0.02180t + M. 37 ...
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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 University of California, Berkeley.

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