Exponential Growth Equation/Special Case

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

Proof
Let $f \left({x}\right) = C e^x$.

Then by Derivative of Exponential Function:


 * $f' \left({x}\right) = f \left({x}\right)$

From Exponential of Zero:


 * $f \left({0}\right) = C$

Hence $C e^x$ is a solution of $y' = y$.

Now suppose that a function $f$ satisfies $f' \left({x}\right) = f \left({x}\right)$.

Consider $h \left({x}\right) = f \left({x}\right) e^{-x}$.

By the Product Rule for Derivatives:

From Zero Derivative implies Constant Function, $h$ must be a constant function.

Therefore, $h \left({x}\right) = h \left({0}\right) = f \left({0}\right)$.

Recalling the definition of $h$, it follows that:


 * $f \left({x}\right) = f \left({0}\right) e^x$