Exponential Growth Equation/Special Case

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Theorem

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

Proof

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

Then by Derivative of Exponential Function:

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

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:

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

$\blacksquare$

Sources

• 1967: Tom M. Apostol: Calculus Volume 1: $\S 8.1$