Differential Equation, Mixture Problem.

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min. How much salt is in the tank (a) after t minutes and (b) after one hour?

Setting our variables: S (or S(t)) = kg of salt at time t, t = time in minutes.

We need to setup our D.E. and solve for S(t) to find our salt concentration for any time, t.

We set up our differential equation structure by saying dS/dt = (rate in) - (rate out). The rate is kg of salt per min.
Finding rate in is [(.05)(5)+(.04)(10)]. Notice liters cancel out to get kg/min.
Finding rate out is (S/1000)(15). S/1000 is the amount of total salt in the mixture. We use S because this changes with time. Again, notice liters cancel out to get our desired rate.
Combining our rate in with our rate out.
Combining the terms so we can multiple/divide with a common denominator.
Combining  "like terms" on each side of the equation and setting up the integral.
Integrating each side.
Need to find the constant, C. We can use our initial condition of S(0) = 0 because at time = 0, we had 0 kg of salt.
Plugging in 0 for S and t.
Solve for C = -(1/3)ln(130)
Putting C into our equation
Multiplying each side by (-3) and then raising each side to the e power.
Remember that E^ln(x) = x, and that the addition of a power of e is the same as multiplying each addition term by e^(each term)
Again e^ln(x) = x. Also, we can remove the absolute value signs as our constraints of the problem require that the value to always be positive.
Solving for S, we get s(t)=-(130/3)e^(-3t/200)+(130/3). This is our equation for the amount of salt for time t.
Solving for S(60), we get 25.72 kg of salt in one hour.