Finding the particular solution of $T(t) = T_a + Ce^{-kt}$ the following is the Newton’s Law of Cooling, 
$$\frac{dT}{dt} = - k (T - T_a) $$
where T(t) is the temperature of object at time t, $T_a$ is the ambient temperature at 25 degrees Cc and k is the positive constant.
From that differential equation, I integrate it to get the general solution of -
$$ T(t) = T_a + Ce^{-kt} = 25 + Ce^{-kt} $$  
I conducted an experiment of boiling water cooling down in room temperature of 25.
At 0min, it’s 100 Degrees C, 
At 1 min, it’s 96
At 5 mins, it’s 88 
using this data, I am trying to find the particular solution, but to do that, I need to find the constant $k$ before I go on to find the arbitrary constant, $C$ .  I am not sure on how to go about that. Thanks.. 
 A: You can linearize the equation:  $T ( t ) = 25 + Ce^{-kt}$
$T  -25= Ce^{-kt}$
$\ln(T-25)=\ln(C)-kt$
This is equivalent to
$y=a-kt$
This is a linear equation. Therefore it is possible to apply $\texttt{linear regression}$
Here you have to evaluate the parameters $a$ and $k$.
For this purpose you have to transform some values, which can be done easily with excel.
$$\begin{array}{|c|c|c|} \hline t&T&\ln(T-25) \\ \hline 0&100&4.317488 \\ \hline1&96&4.262679 \\  \hline5&88&4.143135 \\  \hline \vdots&\vdots&\vdots \end{array}$$
Remark
Linear regression can be done in Excel with matrices. The formula is
$\textbf{b}=(\textbf{X} '\textbf{X})^{-1}\textbf X'\textbf{y}$. For more detailed information look here, for instance. More specific:
$\textbf{b}=\begin{pmatrix} a\\ -k \end{pmatrix}, \quad\textbf{X}=\begin{pmatrix}{} 1 &0 \\ 1 & 1\\ 1&5\\ \vdots & \vdots \end{pmatrix}\quad \textbf{y}=\begin{pmatrix} 4.317488\\ 4.262679\\ 4.143135\\ \vdots \end{pmatrix}$
A: Hint.
Consider the new variable $\Theta = T-T_0$ so the new DE reads
$$
\frac{d\Theta}{dt} = -k\Theta
$$
A: Take the first 2 set of data substitute them in your equation and you will get
$$k=ln((T 1-Ta)/(T 2-Ta)) / (t 2-t 1))$$
Then use $k$ to calculate independently $C$ in terms of the 3rd set of data.
