# Exponential Growth Equation/Special Case

 It has been suggested that this page or section be merged into First Order ODE/dy = k y dx. (Discuss)

## Theorem

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

## Proof

Let $\map f x = C e^x$.

$\map {f'} x = \map f x$

From Exponential of Zero:

$\map f 0 = C$

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

Now suppose that a function $f$ satisfies $\map {f'} x = \map f x$.

Consider $\map h x = \map f x e^{-x}$.

By the Product Rule for Derivatives:

 $\ds \map {h'} x$ $=$ $\ds \map {f'} x e^{-x} - \map f x e^{-x}$ $\ds$ $=$ $\ds \map f x e^{-x} - \map f x e^{-x}$ $\ds$ $=$ $\ds 0$

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

Therefore, $\map h x = \map h 0 = \map f 0$.

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

$\map f x = \map f 0 e^x$

$\blacksquare$