# Differential equation involving rate in - rate out

I have been struggling with this kind of problem ever since I can analyze the rate-in and rate-out with other brine problems. It is fine for me to show up the differential equation beyond your solution.

A tank contains 400 liters of brine. Twelve (12) liters of brine, each containing 2.50 kg of dissolved salt enter the tank per minute and the mixture kept uniform by stirring, leaves at the rate of 8 liters per minute. If the concentration is to be 2.00 kg per liter at the end of one hour, how many kgs of salt were originally present in the tank?

As what I have understood, my rate-in would probably 2.50 kg/min which is the given. However, I don't know much about the rate-out because of how to formulate the concentration of the tank.

Let $y(t)$ be the kg's of salt in the tank at time $t$, and let $A$ be the kgs of salt initially in the tank. We need the change in $y(t)$.
I think it's best to use "(rate-in)(concentration in) - (rate-out)(concentration out)." Your rate in is $12$ l/m and the concentration is $2.5$ kg/l. So the amount of salt entering the tank is $12\cdot 2.5 = 30$. The rate out is $8$ l/m. The concentration out is trickier. It equals the amount of sale in the tank at time $t$ (call it $y(t)$) divided by the amount of water, which will be $400+4t$ because we start with $400$ liters and have a net increase of $4 (=12-8)$ liters every minute. So the concentration out is $y/(400+4t)$.
The to write the IVP, we set $y'$ equal to all the changes in $y$:
$$y' = 30 - \frac{y}{400+4t} \; y(0) =A.$$
This is a linear IVP, which is straightforward to solve, but $A$ will be involved in the solution. Plug $200$ in for $y$ and $60 (= 1$ hour) for $t$ and solve for $A$.