A solution containing 90% by volume of alcohol (in water) runs at 1 gal/min into a 100-gal tank of pure water where it is continually mixed. The mixture is withdrawn at the rate of 1 gal/min. When will it start coming out 50% alcohol?

The given tank of water contains 100 gal/min of pure water and 0.9 gal/min of alcohol. The mixture of water and alcohol is withdrawn at the rate of 1 gal/min.

An ordinary differential equation (ODE) is a differential equation containing one or even more functions of one independent variable and derivatives.

Let $f\left(t\right)$be the function that indicates the amount of the alcohol in the mixture at any time.

$\frac{df}{dt}-0.9-\frac{f}{100}$

Solve the above differential equation using the variable separable method.

$$\begin{gathered} \int \frac{df}{90-f}=\int \frac{dt}{100} \\ -In\left(90-f\right)=\frac{t}{100}+C \end{gathered}$$

The amount of alcohol present in the tank at any time is given below.

$f\left(t\right)=90-C_1{e}^{\frac{-1}{100}}$
For solving, use the initial condition such as at a time $t=0$alcohol inside the tank is zero.

Thus, role="math" localid="1655199566724" $90-C_1\Rightarrow C_1=90$.

The amount of alcohol present in the tank is half of its volume.

role="math" localid="1655199656094" $$\begin{gathered} 50=90-90e^{\frac{-t}{100}} \\ 50=90-90e^{\frac{-t}{100}} \\ 50=90\left(1-e^{\frac{-t}{100}}\right) \end{gathered}$$

Simplify further

$$\begin{gathered} \frac{5}{9}=\left(1-e^{\frac{-t}{100}}\right) \\ 1-\frac{5}{9}=e^{\frac{-t}{100}} \\ \frac{4}{9}=e^{\frac{-t}{100}} \end{gathered}$$

Simplify further

$$\begin{gathered} t=-100In\frac{4}{9} \\ t=81.1 \end{gathered}$$

Therefore, the time when the mixture in the tank contains 50% of alcohol of its total volume is 81.1 minutes.