Exponential Growth Equation/Special Case

Theorem
All solutions of the differential equation $y' = y$ take the form $y=Ce^x$.

This can also be expressed as an existence-uniqueness theorem. The unique function $f$ satisfying $f'(x)=f(x)$ for all $x \in \R$ and $f(0)=C$ is the function $f(x)=Ce^x$.

Proof
It is clear that $f(x)=Ce^x$ satisfies the equation $f'(x)=f(x)$ and $f(0)=C$ by Derivative of the Exponential Function.

Now suppose that a function $f$ satisfies $f'(x) = f(x)$. Consider $h(x) = f(x)e^{-x}$. Then, by the Product Rule,

$ \begin{align*} h'(x) &= f'(x)e^{-x} - f(x)e^{-x} \\ &= f(x)e^{-x} - f(x)e^{-x} \\ &= 0 \end{align*} $

$h$ has a $0$ derivative, so it must be a constant function.

Therefore, $h(x) = h(0) = f(0)$. From this we can conclude that $f(x) = f(0)e^x$.

Source

 * : $\S 8.1$