4.3 Separable equations  (Page 3/8)

The ambient temperature (surrounding temperature) is $75\text{°}\text{F},$ so $T_s=75.$ The temperature of the pizza when it comes out of the oven is $350\text{°}\text{F},$ which is the initial temperature (i.e., initial value), so $T_0=350.$ Therefore [link] 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 $T=75$ as a constant solution. Since the pizza starts at $350\text{°}\text{F},$ this is not the solution we are seeking.
2. Rewrite the differential equation by multiplying both sides by $dt$ and dividing both sides by $T-75\text{:}$
   
   $\frac{dT}{T-75}=kdt.$
3. Integrate both sides:
   
   $\begin{array}{ccc}\hfill {\displaystyle \int \frac{dT}{T-75}}& =\hfill & {\displaystyle \int kdt}\hfill \\ \hfill \text{ln}\left|T-75\right|& =\hfill & kt+C.\hfill \end{array}$
4. Solve for $T$ by first exponentiating both sides:
   
   $\begin{array}{ccc}\hfill e^{\text{ln}\left|T-75\right|}& =\hfill & e^{kt+C}\hfill \\ \hfill \left|T-75\right|& =\hfill & C_1{e}^{kt}\hfill \\ \hfill T-75& =\hfill & C_1{e}^{kt}\hfill \\ \hfill T(t)& =\hfill & 75+C_1{e}^{kt}.\hfill \end{array}$
5. Solve for $C_1$ by using the initial condition $T\left(0\right)=350\text{:}$
   
   $\begin{array}{ccc}\hfill T(t)& =\hfill & 75+C_1{e}^{kt}\hfill \\ \hfill T\left(0\right)& =\hfill & 75+C_1{e}^{k\left(0\right)}\hfill \\ \hfill 350& =\hfill & 75+C_1\hfill \\ \hfill C_1& =\hfill & 275.\hfill \end{array}$
   
   Therefore the solution to the initial-value problem is
   
   $T(t)=75+275e^{kt}.$
   
   To determine the value of $k,$ we need to use the fact that after $5$ minutes the temperature of the pizza is $340\text{°}\text{F}\text{.}$ Therefore $T\left(5\right)=340.$ Substituting this information into the solution to the initial-value problem, we have
   
   $\begin{array}{ccc}\hfill T(t)& =\hfill & 75+275e^{kt}\hfill \\ \hfill T\left(5\right)& =\hfill & 340=75+275e^{5k}\hfill \\ \hfill 265& =\hfill & 275e^{5k}\hfill \\ \hfill e^{5k}& =\hfill & \frac{53}{55}\hfill \\ \hfill \text{ln}{e}^{5k}& =\hfill & \text{ln}\left(\frac{53}{55}\right)\hfill \\ \hfill 5k& =\hfill & \text{ln}\left(\frac{53}{55}\right)\hfill \\ \hfill k& =\hfill & \frac{1}{5}\text{ln}\left(\frac{53}{55}\right)\approx -\mathrm{0.007408.}\hfill \end{array}$
   
   So now we have $T(t)=75+275e^{\mathrm{-0.007048}t}.$ When is the temperature $300\text{°}\text{F?}$ Solving for $t,$ we find
   
   $\begin{array}{ccc}\hfill T(t)& =\hfill & 75+275e^{\mathrm{-0.007048}t}\hfill \\ \hfill 300& =\hfill & 75+275e^{\mathrm{-0.007048}t}\hfill \\ \hfill 225& =\hfill & 275e^{\mathrm{-0.007048}t}\hfill \\ \hfill e^{\mathrm{-0.007048}t}& =\hfill & \frac{9}{11}\hfill \\ \hfill \text{ln}{e}^{\mathrm{-0.007048}t}& =\hfill & \text{ln}\frac{9}{11}\hfill \\ \hfill \mathrm{-0.007048}t& =\hfill & \text{ln}\frac{9}{11}\hfill \\ \hfill t& =\hfill & -\frac{1}{0.007048}\text{ln}\frac{9}{11}\approx \mathrm{28.5.}\hfill \end{array}$
   
   Therefore we need to wait an additional $23.5$ minutes (after the temperature of the pizza reached $340\text{°}\text{F}).$ That should be just enough time to finish this calculation.