A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drains from the tank at the rate 7 L/min. Let y o u be the number of kg of salt in the tank after t minutes. The differential equation for this situation would be:

dy/dt = 7y / (t - 1000)

Step-by-step explanation:

Change in mass of salt = mass of salt going in - mass of salt going out

dy/dt = 0 - (C kg/L * 7 L/min)

where C is the concentration of salt in the tank.

The concentration is mass divided by volume:

C = y / V

The volume in the tank as a function of time is:

V = 1000 + 6t - 7t

V = 1000 - t

Therefore:

C = y / (1000 - t)

Substituting:

dy/dt = - 7y / (1000 - t)

dy/dt = 7y / (t - 1000)

If we wanted, we could separate the variables and integrate. But the problem only asks that we find the differential equation, so here's the answer.