17. Applications of Differential Equations

Balancing problems occur when there is some type of quantity to which you are adding and subtracting at specified rates. The solution begins by constructing the differential equation for the rate of change of the quantity, balancing the rate in minus the rate out.

c. Mixing Problems

Mixing problems are a special case of balancing problems when a material is mixed into a solution.

Initially, \(30\) cups of sugar are put into a \(20\) liter vat of boiling water. Pure water boils away at the rate of \(2\) liters per hour. As it boils, more sugar water containing \(2\) cups of sugar per liter is added to the vat at \(3\) liters per hour. The sugar water is kept thoroughly mixed and removed from the vat through a spout at the bottom at \(1\) liter per hour. How much sugar is in the vat at time \(t\)? How much sugar is in the vat after \(6\) hours? What is the concentration of sugar (in cups per liter) in the water coming out the spout after \(6\) hours?

First notice that each hour, \(2\) liters boil away while \(1\) liter is removed through the spout. In the same time \(3\) liters are added. So there are always \(20\) liters of sugar water in the vat.
To find the quantity \(S(t)\) of sugar in the vat, we first need to construct the differential equation for its rate of change \(\dfrac{dS}{dt}\). This equation has the form: \[ \dfrac{dS}{dt} =\left[\text{rate sugar added}\right]-\left[\text{rate sugar removed}\right] \] Since \(S(t)\) is measured in cups of sugar, while \(t\) is measured in hours, \(\dfrac{dS}{dt}\) is measured in cups per hour, or \(\dfrac{\text{cups}}{\text{hour}}\). So the rate sugar is added and removed must also be in \(\dfrac{\text{cups}}{\text{hour}}\). First, sugar water is added at the rate of \(3\,\dfrac{\text{liters}}{\text{hour}}\) and this water contains \(2\,\dfrac{\text{cups}}{\text{liter}}\). So sugar is being added at the rate of \[ \text{rate sugar added} =2\,\dfrac{\text{cups}}{\text{liter}}\cdot3\,\dfrac{\text{liters}}{\text{hour}} =6\,\dfrac{\text{cups}}{\text{hour}} \] Notice how the units cancel. Second sugar water is removed at the rate of \(1\,\dfrac{\text{liter}}{\text{hour}}\). To know the rate in \(\dfrac{\text{cups}}{\text{hour}}\), we need to know the concentration in \(\dfrac{\text{cups}}{\text{liter}}\). At time \(t\), there are \(S(t)\) cups of sugar and \(20\) liters of water in the vat. So the concentration in the vat is \(\dfrac{S(t)}{20}\,\dfrac{\text{cups}}{\text{liter}}\) and the rate sugar is coming out the spout is \[ \text{rate sugar removed} =\dfrac{S(t)}{20}\,\dfrac{\text{cups}}{\text{liter}}\cdot1\,\dfrac{\text{liter}}{\text{hour}} =\dfrac{S(t)}{20}\,\dfrac{\text{cups}}{\text{hour}} \] Finally, pure water boils away at the rate of \(2\,\dfrac{\text{liters}}{\text{hour}}\). However, since pure water contains no sugar, there is no effect on the equation. So the differential equation becomes: \[ \dfrac{dS}{dt}=6-\dfrac{S(t)}{20} \] This equation is both separable and linear. We choose to solve it as a linear equation. In standard form it is \[ \dfrac{dS}{dt}+\dfrac{1}{20}S=6 \] We compute the integrating factor: \[ I=e^{\textstyle \int \frac{1}{20}\,dt}=e^{t/20} \] Then we multiply the standard equation by the integrating factor and integrate: \[ \dfrac{d}{dt}\left(e^{t/20}S\right) =e^{t/20}\dfrac{dS}{dt}+e^{t/20}\dfrac{1}{20}S=6e^{t/20} \] \[ e^{t/20}S=\int 6e^{t/20}\,dt=120e^{t/20}+C \] Finally we solve for \(S\) by multiplying by \(e^{-\,t/20}\): \[ S=120+Ce^{-t/20} \] To find \(C\), we use the initial condition \(S(0)=30\): \[ S(0)=120+Ce^{-0/20}=120+C=30 \] So \(C=-90\) and the solution is \[ S(t)=120-90e^{-t/20} \] After \(6\) hours, there are \[ S(6)=120-90e^{-6/20}\approx53.3\,\text{cups} \] of sugar in the vat and the concentration is \[ \dfrac{53.3}{20}\,\dfrac{\text{cups}}{\text{liter}} =2.665\,\dfrac{\text{cups}}{\text{liter}} \]