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:

\(\displaystyle h' \left({x}\right)\)

\(=\)

\(\displaystyle f' \left({x}\right) e^{-x} - f \left({x}\right) e^{-x}\)

\(\displaystyle \)

\(=\)

\(\displaystyle f \left({x}\right) e^{-x} - f \left({x}\right) e^{-x}\)

\(\displaystyle \)

\(=\)

\(\displaystyle 0\)

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:

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

$\blacksquare$

Sources

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