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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.
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 $k$
