Differential equation $f'(x)=\alpha\cdot f(x-1)^\beta$

Asked Jul 21 '15 at 15:25

Active Jul 21 '15 at 15:30

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Is there a way to solve $$f'(x)=\alpha\cdot f(x-1)^\beta,$$ where $\alpha>0,\beta\neq0.$

I know that if the arguments matched, I could use separation of variables to get $\frac{1}{1-\beta}f(x)^{1-\beta}=ax+C$, unless $\beta=1$, in which case I get a log. Here though, I don't even know how to start because the derivative is in terms of $x$ while the function itself is in terms of $x-1$.

edited Jul 21 '15 at 15:30

hardmath

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asked Jul 21 '15 at 15:25

Albert Fletcher

1

This is an example of a delay differential equation because the derivative of $f$ is related to values of the function at previous times (just one previous point in time in this case). – hardmath Jul 21 '15 at 15:30
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At las many solutions exist on left-bounded intervals. For example, define $f$ arbitryrily (continuous of course) on $[0,1]$ with the sole restriction that $f'(1)$ exists and is $=\alpha f(0)^\beta$. Then you can extend to $[0,\infty)$ successively by integrating – Hagen von Eitzen Jul 21 '15 at 15:31
You can see this techniques of continuation here – Michael Galuza Jul 21 '15 at 15:35

Differential equation $f'(x)=\alpha\cdot f(x-1)^\beta$

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