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Unformatted text preview: Q ( t ) = inow  outow = (0 . 2)(2)(2)( Q 200 ) or Q + Q 100 = 0 . 4 . Integrating factor is = exp( Z dt 100 ) = exp( t 100 ) . Then d dt (exp( t 100 ) Q ) = Z . 4 exp( t 100 ) dt. This gives exp( t 100 ) Q = . 4 1 / 100 exp( t 100 ) + c and thus Q ( t ) = 40 + c exp(t 100 ) . Initial condition Q (0) = 0 gives c =40 so the nal answer is Q ( t ) = 40(1exp(t 100 )) . 2. Find the amount of salt in the tank at the end of 50 minutes. Answer: Calculate Q (50) = 40(1exp(50 / 100)) = 15 . 7387761. 3. If the pumping is continued indenitely, what is the limiting amount of salt? Answer: The limit is clearly 40(10) = 40. 4. When will the resulting amount of salt equal half the limiting amount found above? Answer: Solve Q ( t ) = 20 which gives 1 2 = 1exp(t 100 ) . It gives t = 100 ln(2) = 69 . 31471806. 2...
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 Fall '09
 EdrayGoins

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