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

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*: $\S 8.1$