How find this function $f(x)=ae^{ex}$ other solution?

Question

let $f(x)$ such $$f'(x)=f\left(x+\dfrac{1}{e}\right)$$

I found this problem one solution(maybe have other)

Let $f(x)=ae^{bx}$ then we have $$abe^{bx}=ae^{bx+\dfrac{b}{e}}\Longrightarrow ae^{bx}(b-e^{b/e})\Longrightarrow b=e$$ so $$y=ae^{ex}$$ is one solution? and How find other solution? Thank you

I wouldn't expect there to be any more solutions. It's a first order (delay) ODE after all, and you've already found a suitable one-parameter family of solutions. — David H, Nov 09 '13 at 15:29
Yes,This is ODE,I think this is one solution.if no,How can prove this is only solution,Thank you — math110, Nov 09 '13 at 15:37
Possible duplicate of Differential equations that are also functional. — Hu Zhengtang, Nov 11 '13 at 16:39

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

1

See here. Yes, this is an answer. You should find more or less everything you need in the linked thread.

edited Apr 13 '17 at 12:20

Community

1

answered Nov 12 '13 at 15:30

Ahaan S. Rungta

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How find this function $f(x)=ae^{ex}$ other solution?

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