Finding the form of all functions $f(t)$ that satisfies $f'(t) = k \cdot f(t)$ I do not know anything about differential equations.
From my textbook, it is shown that $f(t) = e^{kt}$ satisfies $f'(t) = k \cdot f(t)$. Which makes perfect sense. But then, it continues by defining a function $g(t) = e^{-kt} \cdot f(t)$.
Continuing...
\begin{align}
      g'(t) &= 0 \;\;\;\; \text{(which shows that $g$ is a constant function)}\\        
      g(t) &= e^{-kt} \cdot f(t) \\
      g(t) \cdot e^{kt} &= f(t) \\
      C \cdot e^{kt} &= f(t) \;\;\;\; \text{(where C is some constant)}\\
      f(t) &= f(0) \cdot e^{kt} \;\;\;\; \text{(I'm assuming $C = f(0)$ because $f(0) = C \cdot e^{k(0)} = C(1) = C$)}
\end{align}
This entire process makes sense to me, but the one question I have is why exactly is $g$ defined as specifically $e^{-kt} \cdot f(t)$? Is it just some function that's specifically defined with the sole purpose of showing the general form of all functions $f$ is $f(0)(e^{kt})$?
 A: Perhaps it is much easier to see it like this: Let $y=f(t)$, then we have
\begin{align*}
y'&=k\cdot y\\[1mm]
\frac{dy}{dt}&=k\cdot y\\[1mm]
\frac{dy}{y}&=k\cdot dt\\[1mm]
\color{red}\int\frac{dy}{y}&=k\cdot \color{red}\int dt\\
\ln(y)&=kt+const.\\[1mm]
y&=e^{const.}\cdot e^{kt}\\[1mm]
y&=\color{red}C e^{kt}
\end{align*}
Then you get the general solution $f(t)=C e^{kt}$, when $C=f(0)$.
But to try to give an idea of what the author has in mind, it may be that you want to introduce, by way of example, the idea of "integrating factor". We have
\begin{align*}
y'&=k\cdot y\\
\color{red}{e^{-kt}}\cdot y'&=k\cdot \color{red}{e^{-kt}}\cdot y\\
\color{red}{e^{-kt}}\cdot y'-k\cdot \color{red}{e^{-kt}}\cdot y&=0\\
\color{red}{e^{-kt}}\cdot y'+\left(\color{red}{e^{-kt}}\right)'\cdot y&=0\\
\left(\color{red}{e^{-kt}}y\right)'&=0
\end{align*}
Thus, if we call $g(t)=e^{-kt}y$ you get $g'(t)=0$. This is the reason why the author takes that form for g, later on you will see a general way of how to obtain these integrating factors.
