A $200$ liter tank contains $100$ liters of pure water. Starting at time $t = 0$, alcohol is pumped into the tank at a rate of two liters per minute. At the same time the well-stirred mixture is pumped out of the tank at a rate of only one liter per minute. $\\$
Derive the initial value problem for $y(t)$ where $y(t)$ denotes the amount of alcohol in the tank at time $t$(minutes) and solve the IVP, in particularly determine the amount of alcohol in the tank when the tank is full.
My progress: I am struggling with construction of these type problems. If I know the differential equation, there is no doubt on solving the DE. My main obstacle is how to understand these problems in terms of rate of changes. Any help will be welcomed!
Best Answer
Put a name on things, introduce variables. It always helps.
Let $T(t)$ be the total amount of liquid in the tank at any time $t$. $2\ L$ coming and $1\ L$ going per minute means $$\frac{dT}{dt} = 1 \ L/min$$ $$\implies T(t) = t+100$$
Here comes the tricky part: forming the DE for $y(t)$.
Notice that every minute $2\ L$ of alcohol is poured, and $1 \ L$ of the solution gets thrown out. At any time, the amount of alcohol in one litre of the solution will be $\frac{y(t)}{T(t)}$ and so we get
$$\frac{dy}{dt} = 2-\frac{y}{T} = 2-\frac{y}{t+100}$$ $$\implies y(t) = \frac{t(t+200)}{t+100}$$