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#Calculus 2 differential equations license
Want to cite, share, or modify this book? This book uses theĬreative Commons Attribution-NonCommercial-ShareAlike License The initial-value problem to be solved is Therefore the differential equation becomes d u d t = 1 − u 50, d u d t = 1 − u 50, and the initial condition is u ( 0 ) = 4. 2 = u ( t ) 50 kg/min, and OUTFLOW RATE is equal to u ( t ) 50.Therefore salt leaves the tank at a rate of u ( t ) 100 Thus, the concentration of salt is u ( t ) 100 u ( t ) 100 kg/L, and the solution leaves the tank at a rate of 2 2 L/min.
The number of kilograms of salt in the tank at time t t is equal to u ( t ). However, the volume of the solution remains fixed at 100 liters. Since the actual amount of salt varies over time, so does the concentration of salt. To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Because solution enters the tank at a rate of 2 2 L/min, and each liter of solution contains 0.5 0.5 kilogram of salt, every minute 2 ( 0.5 ) = 1 kilogram 2 ( 0.5 ) = 1 kilogram of salt enters the tank. INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank.
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