Unformatted text preview: − 5 = 1 liter per minute. Thus, for t ≥ 10, the volume of the solution in the tank is 500 − 1 · ( t − 10) = 510 − t liters. This gives the concentration of salt in the tank A ( t ) 510 − t kg / L (2.11) and rate of exit = 6 L / min × A ( t ) 510 − t kg / L = 6 A ( t ) 510 − t kg / min . Hence, the di±erential equation, for t > 10, becomes dA dt = 1 − 6 A 510 − t ⇒ dA dt + 6 A 510 − t = 1 with the initial condition A (10) = 14 . 04 (the value found in (a) ). This equation is a linear equation. We have µ ( t ) = exp ±Z 6 510 − t dt ² = exp ( − 6 ln  510 − t  ) = (510 − t ) − 6 55...
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, International System of Units, Kilogram, Litre, initial condition, Cubic metre

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