108_pdfsam_math 54 differential equation solutions odd

108_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 and so the mass of salt in the tank after t minutes is x ( t )=5 4 . 5 e 2 t/ 25 . If the concentration of salt in the tank is 0 . 02 kg/L, then the mass of salt is 0 . 02 × 100 = 2 kg, and, to Fnd this moment, we solve 5 4 . 5 e 2 t/ 25 =2 e 2 t/ 25 = 2 3 t = 25 ln(3 / 2) 2 5 . 07 (min) . 3. Let x ( t ) be the volume of nitric acid in the tank at time t . The tank initially held 200 L of a 0 . 5% nitric acid solution; therefore, x (0) = 200 × 0 . 005 = 1. Since 6 L of 20% nitric acid solution are flowing into the tank per minute, the rate at which nitric acid is entering is 6 × 0 . 2=1 . 2 L/min. Because the rate of flow out of the tank is 8 L/min and the rate of flow in is only 6 L/min, there is a net loss in the tank of 2 L of solution every minute. Thus, at any time t , the tank will be holding 200 2 t liters of solution. Combining this
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