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Unformatted text preview: tank and 3 gallons leave per minute, the number of gallons of water in the tank is (100-t ).) The rate at which water is leaving the tank is 3 gallons per minute, so the rate of salt being drained out is simply 3 x 100-t . We now have dx dt = 1-3 x 100-t dx dt + 3 x 100-t = 1 This is a linear diﬀerential equation, which can be solved by using the integrating factor e R 3 100-t dt = e-3log | 100-t | = (100-t )-3 : (100-t )-3 dx dt + 3 x (100-t )-4 = (100-t )-3 d dt ( x (100-t )-3 ) = (100-t )-3 x (100-t )-3 = 1 2 (100-t )-2 + C x = 1 2 (100-t ) + C (100-t ) 3 Since the tank contains no salt at the beginning, we have the initial condition x (0) = 0. Therefore we have 0 = 50 + C (100) 3 , so C =-50 100 3 =-1 20000 and 1 x ( t ) = 1 2 (100-t )-1 20000 (100-t ) 3 . (Note: at t = 100 minutes, the tank is empty, so our diﬀerential equation no longer applies. It is not necessary to note this for the solution though.) 2...
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This note was uploaded on 04/02/2008 for the course MATH 33B taught by Professor Staff during the Spring '07 term at UCLA.
- Spring '07