First we define a function [latex]u\left(t\right)[/latex] that represents the amount of salt in kilograms in the tank as a function of time. Then [latex]\frac{du}{dt}[/latex] represents the rate at which the amount of salt in the tank changes as a function of time. Also, [latex]u\left(0\right)[/latex] represents the amount of salt in the tank at time [latex]t=0[/latex], which is [latex]4[/latex] kilograms.

The general setup for the differential equation we will solve is of the form

INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of [latex]2[/latex] L/min, and each liter of solution contains [latex]0.5[/latex] kilogram of salt, every minute [latex]2\left(0.5\right)=1\text{kilogram}[/latex] of salt enters the tank. Therefore INFLOW RATE = [latex]1[/latex].

To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time [latex]t[/latex] is equal to [latex]u\left(t\right)[/latex]. Thus, the concentration of salt is [latex]\frac{u\left(t\right)}{100}[/latex] kg/L, and the solution leaves the tank at a rate of [latex]2[/latex] L/min. Therefore salt leaves the tank at a rate of [latex]\frac{u\left(t\right)}{100}\cdot 2=\frac{u\left(t\right)}{50}[/latex] kg/min, and OUTFLOW RATE is equal to [latex]\frac{u\left(t\right)}{50}[/latex]. Therefore the differential equation becomes [latex]\frac{du}{dt}=1-\frac{u}{50}[/latex], and the initial condition is [latex]u\left(0\right)=4[/latex]. The initial-value problem to be solved is

The differential equation is a separable equation, so we can apply the five-step strategy for solution.

Step 1. Setting [latex]1-\frac{u}{50}=0[/latex] gives [latex]u=50[/latex] as a constant solution. Since the initial amount of salt in the tank is [latex]4[/latex] kilograms, this solution does not apply.

Step 2. Rewrite the equation as

Then multiply both sides by [latex]dt[/latex] and divide both sides by [latex]50-u\text{:}[/latex]

Step 3. Integrate both sides:

Step 4. Solve for [latex]u\left(t\right)\text{:}[/latex]

Eliminate the absolute value by allowing the constant to be either positive or negative:

Finally, solve for [latex]u\left(t\right)\text{:}[/latex]

Step 5. Solve for [latex]{C}_{1}\text{:}[/latex]

The solution to the initial value problem is [latex]u\left(t\right)=50 - 46{e}^{\frac{\text{-}t}{50}}[/latex]. To find the limiting amount of salt in the tank, take the limit as [latex]t[/latex] approaches infinity:

Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is [latex]50[/latex] kilograms, then it remains constant. If it starts at less than 50 kilograms, then it approaches 50 kilograms over time.

The ambient temperature (surrounding temperature) is [latex]75^\circ\text{F}[/latex], so [latex]{T}_{s}=75[/latex]. The temperature of the pizza when it comes out of the oven is [latex]350^\circ\text{F}[/latex], which is the initial temperature (i.e., initial value), so [latex]{T}_{0}=350[/latex]. Therefore our equation becomes

To solve the differential equation, we use the five-step technique for solving separable equations.

1. Setting the right-hand side equal to zero gives [latex]T=75[/latex] as a constant solution. Since the pizza starts at [latex]350^\circ\text{F}[/latex], this is not the solution we are seeking.
2. Rewrite the differential equation by multiplying both sides by [latex]dt[/latex] and dividing both sides by [latex]T - 75\text{:}[/latex]
   
   [latex]\frac{dT}{T - 75}=kdt[/latex].

    

3. Integrate both sides:
   
   [latex]\begin{array}{ccc}\hfill {\displaystyle\int \frac{dT}{T - 75}}& =\hfill & {\displaystyle\int kdt}\hfill \\ \hfill \text{ln}|T - 75|& =\hfill & kt+C.\hfill \end{array}[/latex]

    

4. Solve for [latex]T[/latex] by first exponentiating both sides:
   
   [latex]\begin{array}{ccc}\hfill {e}^{\text{ln}|T - 75|}& =\hfill & {e}^{kt+C}\hfill \\ \hfill |T - 75|& =\hfill & {C}_{1}{e}^{kt}\hfill \\ \hfill T - 75& =\hfill & {C}_{1}{e}^{kt}\hfill \\ \hfill T\left(t\right)& =\hfill & 75+{C}_{1}{e}^{kt}.\hfill \end{array}[/latex]

    

5. Solve for [latex]{C}_{1}[/latex] by using the initial condition [latex]T\left(0\right)=350\text{:}[/latex]
   
   [latex]\begin{array}{ccc}\hfill T\left(t\right)& =\hfill & 75+{C}_{1}{e}^{kt}\hfill \\ \hfill T\left(0\right)& =\hfill & 75+{C}_{1}{e}^{k\left(0\right)}\hfill \\ \hfill 350& =\hfill & 75+{C}_{1}\hfill \\ \hfill {C}_{1}& =\hfill & 275.\hfill \end{array}[/latex]

   
   
   Therefore the solution to the initial-value problem is
   

   [latex]T\left(t\right)=75+275{e}^{kt}[/latex].

   
   
   To determine the value of [latex]k[/latex], we need to use the fact that after [latex]5[/latex] minutes the temperature of the pizza is [latex]340^\circ\text{F}\text{.}[/latex] Therefore [latex]T\left(5\right)=340[/latex]. Substituting this information into the solution to the initial-value problem, we have
   

   [latex]\begin{array}{ccc}\hfill T\left(t\right)& =\hfill & 75+275{e}^{kt}\hfill \\ \hfill T\left(5\right)& =\hfill & 340=75+275{e}^{5k}\hfill \\ \hfill 265& =\hfill & 275{e}^{5k}\hfill \\ \hfill {e}^{5k}& =\hfill & \frac{53}{55}\hfill \\ \hfill \text{ln}{e}^{5k}& =\hfill & \text{ln}\left(\frac{53}{55}\right)\hfill \\ \hfill 5k& =\hfill & \text{ln}\left(\frac{53}{55}\right)\hfill \\ \hfill k& =\hfill & \frac{1}{5}\text{ln}\left(\frac{53}{55}\right)\approx -0.007408.\hfill \end{array}[/latex]

   
   
   So now we have [latex]T\left(t\right)=75+275{e}^{-0.007408t}[/latex]. When is the temperature [latex]300^\circ\text{F?}[/latex] Solving for [latex]t[/latex], we find
   

   [latex]\begin{array}{ccc}\hfill T\left(t\right)& =\hfill & 75+275{e}^{-0.007408t}\hfill \\ \hfill 300& =\hfill & 75+275{e}^{-0.007408t}\hfill \\ \hfill 225& =\hfill & 275{e}^{-0.007408t}\hfill \\ \hfill {e}^{-0.007408t}& =\hfill & \frac{9}{11}\hfill \\ \hfill \text{ln}{e}^{-0.007408t}& =\hfill & \text{ln}\frac{9}{11}\hfill \\ \hfill -0.007408t& =\hfill & \text{ln}\frac{9}{11}\hfill \\ \hfill t& =\hfill & -\frac{1}{0.007408}\text{ln}\frac{9}{11}\approx 27.09.\hfill \end{array}[/latex]

   
   
   Therefore we need to wait an additional [latex]22.09[/latex] minutes (after the temperature of the pizza reached [latex]340^\circ\text{F}[/latex]). That should be just enough time to finish this calculation.