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Consider the equation

$$ f'(t)=\frac{f(t-b)}{t-b}$$ $f'(t)=\frac{df(t)}{dt}$ and $b$ is a constant.

Does anyone know if this equation has a name, an analytic solution and how to find the solution?

This is not a question about how to solve the equation numerically.

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    $\begingroup$ @doraemonpaul Why the tag difference equations? $\endgroup$ – bob Jun 8 '13 at 8:49
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    $\begingroup$ $f(t) = ct$ for some constant $c$ is a solution. Not sure if it's the only one, but you might be able to do something with series to show that it is/isn't. $\endgroup$ – Mr. G Jun 8 '13 at 20:20
  • $\begingroup$ If you know boundary values you can approximate the function. $\endgroup$ – AnilB Jun 8 '13 at 21:44
  • $\begingroup$ @Mr.G +1. I'm too lazy to do it, but expanding $1/(t-b)f(t-b)$ into a Taylor serie around 0 might give you something. since the equation is linear, to prove unicity you'll just have to prove that an analytic solution with $f(0)=0$ has to be $ct$ $\endgroup$ – Glougloubarbaki Jun 8 '13 at 21:48
  • $\begingroup$ See math.stackexchange.com/q/243379/442 and math.stackexchange.com/q/45392/442 for more on delay differential equations. $\endgroup$ – GEdgar Jun 10 '13 at 13:56
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Assume $b\neq0$ to maintain the key meaning of this question:

$f'(t)=\dfrac{f(t-b)}{t-b}$

$(t-b)f'(t)=f(t-b)$

Let $f(t)=\int_p^qe^{t\tau}F(\tau)~d\tau$ ,

Then $(t-b)\int_p^q\tau e^{t\tau}F(\tau)~d\tau=\int_p^qe^{(t-b)\tau}F(\tau)~d\tau$

$t\int_p^q\tau e^{t\tau}F(\tau)~d\tau-b\int_p^q\tau e^{t\tau}F(\tau)~d\tau-\int_p^qe^{-b\tau}e^{t\tau}F(\tau)~d\tau=0$

$\int_p^q\tau e^{t\tau}F(\tau)~d(t\tau)-\int_p^q(b\tau+e^{-b\tau})F(\tau)e^{t\tau}~d\tau=0$

$\int_p^q\tau F(\tau)~d(e^{t\tau})-\int_p^q(b\tau+e^{-b\tau})F(\tau)e^{t\tau}~d\tau=0$

$[\tau F(\tau)e^{t\tau}]_p^q-\int_p^qe^{t\tau}~d(\tau F(\tau))-\int_p^q(b\tau+e^{-b\tau})F(\tau)e^{t\tau}~d\tau=0$

$[\tau F(\tau)e^{t\tau}]_p^q-\int_p^q(\tau F'(\tau)+F(\tau))e^{t\tau}~d\tau-\int_p^q(b\tau+e^{-b\tau})F(\tau)e^{t\tau}~d\tau=0$

$[\tau F(\tau)e^{t\tau}]_p^q-\int_p^q(\tau F'(\tau)+(1+b\tau+e^{-b\tau})F(\tau))e^{t\tau}~d\tau=0$

$\therefore\tau F'(\tau)+(1+b\tau+e^{-b\tau})F(\tau)=0$

$\tau F'(\tau)=-(1+b\tau+e^{-b\tau})F(\tau)$

$\dfrac{F'(\tau)}{F(\tau)}=-\dfrac{1}{\tau}-b-\dfrac{e^{-b\tau}}{\tau}$

$\int\dfrac{F'(\tau)}{F(\tau)}~d\tau=\int\left(-\dfrac{1}{\tau}-b-\dfrac{e^{-b\tau}}{\tau}\right)d\tau$

$\ln F(\tau)=-\ln\tau-b\tau-\int_k^\tau\dfrac{e^{-b\tau}}{\tau}d\tau+c$

$F(\tau)=\dfrac{Ce^{-b\tau-\int_k^\tau\frac{e^{-b\tau}}{\tau}d\tau}}{\tau}$

$\therefore f(t)=\int_p^q\dfrac{Ce^{(t-b)\tau-\int_k^\tau\frac{e^{-b\tau}}{\tau}d\tau}}{\tau}d\tau$ for some suitable constant $p$ , $q$ and $k$

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  • $\begingroup$ Thanks! I think you have a mistake on the sixth line (the integration bounds of the first integral should change with the change of variable). $\endgroup$ – bob Jun 13 '13 at 3:26

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