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Unformatted text preview: β 5 = 1 liter per minute. Thus, for t β₯ 10, the volume of the solution in the tank is 500 β 1 Β· ( t β 10) = 510 β t liters. This gives the concentration of salt in the tank A ( t ) 510 β t kg / L (2.11) and rate of exit = 6 L / min Γ A ( t ) 510 β t kg / L = 6 A ( t ) 510 β t kg / min . Hence, the diΒ±erential equation, for t > 10, becomes dA dt = 1 β 6 A 510 β t β dA dt + 6 A 510 β 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 β t dt Β² = exp ( β 6 ln  510 β t  ) = (510 β t ) β 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.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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