Is there a real-valued function f different from exp($a^x$) for which f'(x) = a*f(x) for all real x?

Question

I would like to ask if the question above is true or not and if true, how it can be proved?

DSolve[f'[x] == a f[x], f[x], x] gives only E^(a x) C[1] as a solution, i.e. a family of exponentials. — corey979, Nov 18 '16 at 12:15
Did you mean for the title to read $\exp(ax)$ instead of $\exp(a^x)$? — zahbaz, Nov 18 '16 at 19:09
Related: How do we show that the function which is its own derivative is exponential?. — dxiv, Nov 18 '16 at 19:24

score 1 · Answer 1 · answered Nov 18 '16 at 13:18

Assume $f: \mathbb{R} \to \mathbb{R}$ is differentiable with $f'(x) = a f(x)$. Consider the function $g(x) = f(x) e^{-ax}$. Then $g$ is differentiable with derivative $$g'(x) = f'(x) e^{-ax} + f(x) (-ae^{-ax}) = e^{-ax}(af(x) - af(x)) = 0.$$

But this means that $g$ is a constant function, i.e. $f(x) = C e^{ax}$.

Is there a real-valued function f different from exp($a^x$) for which f'(x) = a*f(x) for all real x?

1 Answers1