# Proving $k e^{x}$ is the ONLY solution to the differential equation $f(x) = f'(x)$ [duplicate]

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Proof that $\exp(x)$ is the only function for which $f(x) = f&amp;#39;(x)$

The title says it all. How do I prove, that the ONLY solution to the differential equation $f(x) = f'(x)$ is of the form $k e^{x}$ ?

Preferably without making too many assumptions, but proving and arriving at the answer at the same time.

## marked as duplicate by David Mitra, Henry T. Horton, user53153, Clive Newstead, Douglas S. StonesJan 22 '13 at 0:16

• f(x)=0 holds true as well but I'd bet you already thought of that. I guess that is the case where k=0. – kaine Jan 21 '13 at 22:48
• @kaine It's contemplated for $k=0$. – Git Gud Jan 21 '13 at 22:49
• @kaine $f(x)=0$ is a special case of $ke^x$ with $k=0$ – Alan Simonin Jan 21 '13 at 22:49
• You mean $k f(x) = f'(x)$ ? – user42761 Jan 21 '13 at 22:51
• that would be the case if f(x) = $j*e^{kx}$ – kaine Jan 21 '13 at 22:54

First note that, it is not hard to prove that if $g'(x) = 0$, then $g(x) = \text{constant}$.

We have $f'(x) = f(x)$. Since $e^x \neq 0$ for all $x \in \mathbb{R}$, we get that $$e^{-x} f'(x) - e^{-x} f(x) = 0$$

Now recall that the product of two differentiable functions is also differentiable and we have the product rule.

Hence, we have that $$0 = e^{-x} f'(x) - e^{-x} f(x) = e^{-x} \dfrac{df}{dx} + \dfrac{d(e^{-x})}{dx} f = \dfrac{d \left(e^{-x} f(x)\right)}{dx}$$

Hence, $$e^{-x} f(x) = \text{ constant} = k \implies f(x) = ke^x$$

• Thank you for the elegant and nicely explained answer, i appreciate it. – highschool student Jan 21 '13 at 23:29

Suppose $f: \mathbb R \to \mathbb R$ is differentiable and satisfies $f'(x) = f(x)$ for all $x \in \mathbb R$. Then $$\frac{\mathrm d}{\mathrm d x}(f(x)e^{-x}) = e^{-x}(f'(x)-f(x)) = 0,$$ so $f(x)e^{-x} = k$ for a constant $k$.

Suppose that $f(x) = f^{'}(x)$ and consider that $$\frac{d}{dx} f(x)e^{-x} = f^{'}(x)e^{-x} - f(x) e^{-x} = 0$$ and therefore we conclude that $f(x) e^{-x}$ is a constant, which implies the result.