Exponential Growth Equation/Special Case

 It has been suggested that this article or section be renamed: If this is a special case, where is the main case? One may discuss this suggestion on the talk page.

Theorem

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

Proof

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

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

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:

 $\displaystyle h' \left({x}\right)$ $=$ $\displaystyle f' \left({x}\right) e^{-x} - f \left({x}\right) e^{-x}$ $\displaystyle$ $=$ $\displaystyle f \left({x}\right) e^{-x} - f \left({x}\right) e^{-x}$ $\displaystyle$ $=$ $\displaystyle 0$

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:

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

$\blacksquare$