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For my homework assignment I need to solve the following differential equation:

$\dfrac{dT}{dt} = -p(T - T_k)$,

where $T$ stands for the temperature of the pizza, $t$ stands for time and $T_k$ stands for the temperature of the kitchen. They gave me the following information:

  1. The initial temperature of the pizza coming out of the freezer at 8pm is $-18^\circ$C.

  2. The temperature of the kitchen is $20^\circ$C.

  3. $p>0$

First I tried solving the differential equation by substituting $y$ for $T-T_k$ and this gave me the following answer:

$T = T_k + (T(0) - T_k)e^{-pt}$, where I can of course plug in the values of $T(0)$ and $T_k$.

But then I tried solving the differential equation by applying integration from the start, and this gave me the following answer:

$T = ce^{-pt} + T_k$, where I solved $c$ by plugging in $T(0) = -18$ and this gave me a value of $-38$ for $c$.

After this I had to calculate the value of $p$, and I found $p=\log(38) - \log(29)$. For this question it doesn't matter what method I use because they will both give me the same answer $T = T_k - 38 e^{-pt}$, but in the following question they asked me to calculate the temperature of the kitchen in order to defrost the pizza by 9pm. This is where I get confused. When I use my first answer to the differential equation ($T = T_k + (T(0) - T_k)e^{-pt})$ I find a $T_k$ value of $58^\circ$C and when I use my second answer to the differential equation ($T = T_k -38e^{-pt}$) I find a $T_k$ value of $29^\circ$C. Which one (or maybe neither of them) is right and why? What am I doing wrong by solving the differential equation?

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Both solutions to the differential equations are correct and should provide the same results. It is obvious that the kitchen temperature of $29 c$ is not realistic so redo the problem with your second formula and get the same result as you have from the first method

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Notice that "$-38$" contains/hides an instance of $T_k$. If you hide that instance, you can't actually solve for $T_k$. To solve for $T_k$, you need to start with the form that shows all the $T_k$s that are present. Another way of saying this:

"$T = T_k + (T(0) - T_k)\mathrm{e}^{-pt}$" is the generic equation. It applies for any choice of $T(0)$ and $T_k$. When you solved for $c$, you actually solved for "$c$, dependent on $T(0)$ and $T_k$". If you change either $T(0)$ or $T_k$, you get a different $c$. That is, particular constants in solutions depend on the parameters present in the equation, so changing the parameters changes the constants (in general -- a careful choice of parameter change would leave $c$ unchanged).

So either use the generic equation, or remember to update your $c$ when you change the parameters.

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  • $\begingroup$ That makes sense, so my first solution to the differential equation was right because I didn't hide an instance of $T_k$ in that solution? $\endgroup$ – Neri Oct 12 '19 at 19:38
  • $\begingroup$ Both solutions are right -- they're actually the same. One solution makes it easier to change the parameters, so it is more helpful in the next step of the problem. $\endgroup$ – Eric Towers Oct 12 '19 at 19:39
  • $\begingroup$ But could I also use the second solution to calculate the $T_k$ in order to defrost the pizza by 9pm? Because I would need to find a $c$ for the solution to the differential equation, and I can't use the same $c$ as before. $\endgroup$ – Neri Oct 12 '19 at 19:42
  • $\begingroup$ @Neri : Notice that when you find $c$, you solve $T(0) = c \cdot 1 + T_k$ for $c$. We can do this "once and for all": $c = T(0) - T_k$. Then, if someone changes the parameters $T(0)$ or $T_k$, we can immediately update $c$. Alternatively, if someone wants to leave $T(0)$ or $T_k$ unspecified, we can replace $c$ with its form in the parameters, $T(0) - T_k$, and proceed with the parameters visible, instead of hiding in $c$. $\endgroup$ – Eric Towers Oct 12 '19 at 19:46
  • $\begingroup$ Thanks! I understand better now. The only thing I am confused about, what answer do they want when they ask for the solution of the differential equation? I know both answers are the same, but which one is more common to give as an answer to the differential equation? $\endgroup$ – Neri Oct 12 '19 at 20:02

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