59_pdfsam_math 54 differential equation solutions odd

59_pdfsam_math 54 - βˆ’ 5 = 1 liter per minute Thus for t β‰₯ 10 the volume of the solution in the tank is 500 βˆ’ 1 Β t βˆ’ 10 = 510 βˆ’ t liters

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Exercises 2.3 100 A = Ce t/ 100 ( C = ± e C 1 ) A = 100 Ce t/ 100 . The initial condition, A (0) = 5 (initially, there were 5 kg of salt in the tank) implies that 5= A (0) = 100 C C =95 . Substituting this value of C into the solution, we have A ( t ) = 100 95 e t/ 100 . Thus the mass of salt in the tank after 10 min is A (10) = 100 95 e 10 / 100 14 . 04 kg , which gives the concentration 14 . 04 kg / 500 L 0 . 0281 kg / L. (b) After the leak develops, the system satisFes a new di±erential equation. While the rate of input remains the same, 1 kg / min, the rate of exit is now di±erent. Since, every minute, 5 liters of the solution is coming in and 5 + 1 = 6 liters are going out, the volume of the solution in the tank decreases by 6
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Unformatted text preview: βˆ’ 5 = 1 liter per minute. Thus, for t β‰₯ 10, the volume of the solution in the tank is 500 βˆ’ 1 Β· ( t βˆ’ 10) = 510 βˆ’ t liters. This gives the concentration of salt in the tank A ( t ) 510 βˆ’ t kg / L (2.11) and rate of exit = 6 L / min Γ— A ( t ) 510 βˆ’ t kg / L = 6 A ( t ) 510 βˆ’ t kg / min . Hence, the diΒ±erential equation, for t > 10, becomes dA dt = 1 βˆ’ 6 A 510 βˆ’ t β‡’ dA dt + 6 A 510 βˆ’ t = 1 with the initial condition A (10) = 14 . 04 (the value found in (a) ). This equation is a linear equation. We have Β΅ ( t ) = exp Β±Z 6 510 βˆ’ t dt Β² = exp ( βˆ’ 6 ln | 510 βˆ’ t | ) = (510 βˆ’ t ) βˆ’ 6 55...
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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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