Let $x\left(t\right)$ denote the amount of salt in the tank at time $t$, with $x$ in pounds, and $t$ in minutes.

The rate at which salt enters the tank is 40 gal/min times $1\/100$ lbs/gal, or $2\/5$ lb/min.

The rate at which salt leaves the tank is $-\left(\frac{40}{1000}\right)x\=-x\/25$ lbs/min.

The rate of change of salt in the tank, namely, $\frac{\mathrm{dx}}{\mathrm{dt}}$ or $\stackrel{\.}{x}$, is then the difference between the rates at which salt enters and leaves the tank. The units would be lbs/min.

The limiting value of $x\left(t\right)$ is 10 lbs, attained when each of the 1000 gallons of brine in the tank has the concentration of the incoming fluid, namely, $1\/100$ lb/gal. This limiting value is corroborated analytically by letting $t\to \infty$ in $x\left(t\right)\=10\+40{\mathit{\ⅇ}}^{-t\/25}$. In fact, this shows $x\>10$ for all finite $t$, an observation useful in a stepwise solution of the IVP by separation of variables.

The differential equation is easily put into the separated form $\frac{\mathrm{dx}}{\frac{2}{5}-\frac{x}{25}}\=\frac{25 \mathrm{dx}}{10-x}\=\mathrm{dt}$ from which, by antidifferentiation of both sides, it follows that $-25 \ln \(\left|10-x\right|\)\=t\+c$. Dividing through by $-25$ and exponentiating both sides leads to

In the third line, $e^{-c\/25}$ is an arbitrary constant $A$ because $c$ is. The absolute values are removed in line four by recalling that $x\>10$ and introducing the appropriate minus sign on the right.

Finally, imposing the initial condition $x\left(0\right)\=50$ gives $50\=10\+A$ so that $A\=40$ and $x\=10\+40 e^{-t\/25}$.$$

The Context Panel can be used to select the right-hand side, and then used to invoke the Plot Builder to draw a graph akin to the one shown in Figure 5.6.5(a).