59_pdfsam_math 54 differential equation solutions odd

# 59_pdfsam_math 54 differential equation solutions odd - −...

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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 satisfies a new differential equation. While the rate of input remains the same, 1 kg / min, the rate of exit is now different. 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: − 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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