1. The problem statement, all variables and given/known data A tank contains 2860 L of pure water. A solution that contains 0.04 kg of sugar per liter enters a tank at the rate 5 L/min The solution is mixed and drains from the tank at the same rate. Find the amount of sugar in the tank after t minutes. 2. Relevant equations 3. The attempt at a solution Obviously it starts with 0 kg sugar and an initial condition at time t = 0. The rate then would be equal to (the rate entering) - (the concentration exiting). Entering = .04*5 = .2 Exiting = (Y*5)/2860 = Y/572 dY/dt = .2 - Y/527 dY/(.2 - Y/572) = dt integrate both sides -572ln(.2-Y/572) = t + C simplify ln(.2 - Y/572) = -t/572 + C .2 - Y/572 = e^(-t/572+C) -Y/572 = Ce^(-t/572) - .2 Y = Ce^(-t/572) + 114.4 initial value t = 0 and Y = 0 0 = Ce^(-0/572) + 114.4 C = -114.4 so Y = 114.4 - 114.4e^(-t/572) Is this right because I've done it a bunch of times and keep getting the same answer, but it's an online problem and it tells me that it's wrong and I can't figure out what I did incorrectly. Thanks!