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## Homework Statement

This is actually a problem from my physics textbook, but I think it's mostly a mathematical problem, which is why I post it here:

Show that the Langevin equation

1: [tex]\frac{dv}{dt}=-\gamma v+\frac{1}{m} F'(t)[/tex]

is solved by

2: [tex]v-v_0e^{-\gamma\tau}=\sum_{k=0}^{N-1}e^{{-\gamma\tau}(N-k)}\frac{1}{m}\int_0^\tau F'(k\tau+s)ds,[/tex]

where

[tex]t=N\tau.[/tex]

## The Attempt at a Solution

First of all, I'm not sure how

[tex]v-v_0e^{-\gamma\tau}[/tex] can be a solution to a differential equation where the unknown variable is [tex]v[/tex], so I put the term [tex]v_0e^{-\gamma\tau}[/tex] on the right-hand side of equation 2 instead.

Next, I'm trying to take the derivative of equation 2 w.r.t. t:

[tex]-\gamma V_0e^{-\gamma t}+\sum_{k=0}^{N-1}\Big(-\gamma \frac{N-k}{N}e^{-\gamma t\frac{N-k}{N}} \frac{1}{m}\int_0^\tau F'(k\tau+s)ds+e^{-\gamma t\frac{N-k}{N}}\frac{1}{m}F'(\frac{t}{N}(k+1))\Big),[/tex]

where

[tex]F'(\frac{t}{N}(k+1))=\frac{d\int_0^\frac{t}{N} F'(k\tau+s)ds}{dt}.[/tex]

My first question is if this derivative is correct so far. If so, my second question is if I could perhaps be given some clues as how to calculate the sum.