# Analytical solution for a simple exponential differential equation

In general, can I find an analytical solution for an equation of the form

$$\dfrac{dV_{(t)}}{dt} = -\dfrac{V_{(t-1)}}{\tau}$$

where $$t$$ is time, $$V$$ is a variable that changes over time, and $$\tau$$ is a time constant. I think the result should be an exponential decay function, of the form

$$V_0 \cdot e^{-\tau \cdot t}$$

where $$V_0$$ is some initial value for $$V$$.

Also, if anyone could recommend a beginner's book/resource for learning how to solve similar problems I would be most grateful.

Edit: Thanks for LutzL for pointing out this is not an ODE but rather a delay DE.

• This is a delay-differential equation. Usually there is no unique solution, every function on $[-1,0]$ can be extended to a solution over $t\ge 0$. Commented Jul 20, 2019 at 13:44
• Thanks, LutzL. Does this mean my example DDE is possible to solve? Commented Jul 20, 2019 at 14:04

This is a delay-differential equation (DDE). Usually there is no unique solution, every function on $$[−1,0]$$ can be extended to a solution over $$t≥0$$ as $$V(t)=V(0)-\int_{-1}^{t-1}\frac{V(s)}{τ}\,ds.$$
You can of course also look for solutions of the form $$V(t)=ae^{bt}.$$ Inserting you find $$abe^{bt}=-\frac{a}{τ}e^{b(t-1)}\implies bτ=-e^{-b}\iff -\frac1{τ}=be^b$$. This has a real solution only if $$τ>e$$, it can be expressed using the Lambert-W function as $$b=W_0(-\frac1{τ})\in[-1,0]$$ or $$b=W_{-1}(-\frac1{τ})\in(-\infty,-1])$$.
• So I think my solution is $V_0 \cdot e^{- \dfrac{t}{\tau}}$. Is it correct? Commented Jul 22, 2019 at 4:13
• No, this is not a solution. $V_0e^{W_0(-\frac1τ)t}$ and $V_0e^{W_{-1}(-\frac1τ)t}$ are the solutions that are exponential functions. Commented Jul 22, 2019 at 5:15