Mixing differential equation problem We have a big tank of water containing 100 $m^3$ of water and initial pollutant concentration of $0.20 \%$. There is a daily inflow of $2$ $m^3$ of water with polutant concentration of $0.05 \%$ and an equal daily outflow of well mixed water in the tank. How long will it take to reduce the pollutant concentration to $0.1 \% $ ?
Attempt
Let $P(t)$ be the amount of pollutant. We have
$$ \frac{ dP }{dt} = r_{in} - r_{out} $$
Rate in is $ 2 \cdot 0.05 = \frac{ 2 \cdot 5}{100} = \frac{1}{10} $ and Rate out is $ \frac{ P(t) }{100} $, so we have
$$ P' = \frac{1}{10} - \frac{ P }{100} $$
and $P(0)= 0.0020$
Is this the model we are looking for??
 A: Hint
You are close. The fundamental concept is conservation of pollutant mass, since there are no reactions to transform the pollutant. Also, I am assuming by $0.05$% you really mean $0.0005$? You seem to switch the meanings when you describe the initial conditions in your model.
The mass input per unit time is given by:
$$\dot{m}_{in} = 2\times\rho_{water}\times0.0005 = 0.001\rho_{water}$$
Let $P_t$ be the mass fraction of the pollutant in the tank (Mass Pollutant/Mass of Mixture). Lets assume the mass of pollutant does not appreciably affect the density of the fluid, i.e., $\rho_{mixture} \approx \rho_{water}$ The mass leaving the tank is:
$$\dot{m}_{out} = 2\times\rho_{water}\times P_t$$
The rate of change in the mass is given by:
$$\dot{m}_{tank} = \dot{m}_{in} - \dot{m}_{out} = \rho_{water}(0.001-2P_t)$$
Since you want to solve for $P_t$, we need to do a little algebra:
$$\dot{m}_{tank} = V_{tank}\rho_{water}\dot{P}_t$$
This implies:
$$\dot{P}_t := P'= \frac{0.001-2P_t}{V_{tank}}=\frac{1}{1\times10^{5}} - \frac{P_t}{50}$$
Where $P(0)=0.2$ and you want to solve for $P(t)=0.1$
