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: $\S 8.1$