Question

What is the differential equation in Problem 10, if the well-stirred solution is pumped out at a faster rate of 3.5gal/min?

Short answer

The differential equation is $\frac{dA}{dt}+\frac{7}{600-t}A=6$.

Step-by-step solution

Definition:

A differential equation is an equation with one or more derivatives of a function.

Find difference in rate:

Consider that a brine solution is pumped into the tank at a rate of 3gal/min and a well-stirred solution is pumped out at a faster rate of 3.5gal/min.

The concentration of the solution entering is, 2lb/gal.

Hence, the rate at which the solution enters is,

$$\begin{gathered} R_{in}=2\text{lb}/\text{gal}\times 3\text{gal}/\text{min} \\ =6\text{lb}/\text{min} \end{gathered}$$

Thus, the difference in the rate is$$\begin{gathered} (3-3.5)=-0.5\text{gal}/\text{min} \\ =-\frac{1}{2}\text{gal}/\text{min.} \end{gathered}$$

Find differential equation:

By above we can say that afterminutes there are $\left(300-\frac{1}{2}t\right)$gallons of brine in the tank.

So, the rate at which salt is leaving is,

$$\begin{gathered} R_{\text{cutt}}=(3.5\text{gal}/\text{min})\left(\frac{A\left(t\right)}{300-(0.5)t}\text{lb}/\text{gal}\right) \\ =3.5\times \frac{A}{300-\frac{t}{2}}\text{lb}/\text{min} \\ =\frac{7A}{600-t}\text{lb}/\text{min} \end{gathered}$$

Thus, the differential equation for the amount of salt, A(t) in the tank at time t is $\frac{dA}{dt}=6-\frac{7A}{600-t}$$\text{Or}\frac{dA}{dt}+\frac{7}{600-t}A=6.$,