Confused by Langevin Equation Trying to understand the Langevin Equation.  In particular, this passage from a Wikipedia article has me confused (section: "Thermal Noise in an Electrical Resistor"):

$\frac{dU}{dt} =-\frac{U}{RC}+\eta \left( t\right),\;\;$ $\left\langle\eta \left( t\right) \eta \left( t^{\prime }\right)\right\rangle =
 \frac{2k_{B}T}{RC^{2}}\delta \left(t-t^{\prime }\right).$
This equation may be used to determine the correlation function
$\left\langle U\left(t\right) U\left(t^{\prime }\right) \right\rangle
=\left( k_{B}T/C\right) \exp \left( -\left\vert t-t^{\prime }\right\vert /RC\right) \approx 2Rk_{B}T\delta \left( t-t^{\prime}\right),$

Here's my derivation of a rather different result:
$\displaystyle \frac{dU(t)}{dt} =-\frac{U(t)}{RC}+\eta \left( t\right)$
$\displaystyle U(t+t^\prime)\frac{dU(t)}{dt} =-\frac{U(t+t^\prime)U(t)} {RC}+U(t+t^\prime)\eta(t)$
Now concentrate on LHS:
$\displaystyle U(t+t^\prime)\frac{dU(t)}{dt} = \frac{d}{dt}(U(t+t^\prime)U(t))-U(t)\frac{d}{dt}U(t+t^\prime)=
\frac{d}{dt}(U(t+t^\prime)U(t))-U(t)\frac{d}{dt^\prime}U(t+t^\prime)$
$\displaystyle=\frac{d}{dt}(U(t+t^\prime)U(t))-\frac{d}{dt^\prime}(U(t)U(t+t^\prime))$
now do ensemble averaging and assuming system is statistically stationary.  Then the first term on the new LHS vanishes as does the second term on the RHS, leaving:
$\displaystyle -\frac{d}{dt^\prime}\langle U(t+t^\prime)U(t)\rangle = -\frac{\langle U(t+t^\prime)U(t)\rangle} {RC}$
which gives the obviously unphysical
$\displaystyle \langle U(0)U(t^\prime)\rangle=\langle U(0)U(0) \rangle \exp(+t^\prime / RC)$
Can anyone tell me where I went wrong and supply the derivation of the Wikipedia result?
 A: You start by solving the differential equation. It is a first order, linear differential equation with constant coefficients. So the solution of the homogenous system is quite simple:
$U(t) = c\cdot e^{-t/RC}$.
Now we solve the particular  system with variation of the constant $c$, which means we try the
Ansatz $U(t) = c(t)\cdot e^{-t/RC}$. This give the differential equation
$c'(t) = \eta(t)e^{t/RC}$, which can simply integrate to $c(t) = \int\limits_{0}^{t} \eta(\hat{t})e^{\hat{t}/RC}d\hat{t} + U_0$.
So the solutions of the langevin equation is
$U(t) = e^{-t/RC}\left(\int\limits_{0}^{\hat{t}} \eta(t)e^{\hat{t}/RC}d\hat{t} + U_0\right)$.
Now we can calculate all moments. The first moment is pretty strait forward
$\langle U(t)\rangle = \left\langle e^{-t/RC}\left(\int\limits_{0}^{t} \eta(\hat{t})e^{\hat{t}/RC}d\hat{t} + U_0\right) \right\rangle = U_0e^{-t/RC}$.
The second moment needs a little more effort.
$\langle U(t)U(t')\rangle = \left\langle e^{-(t+t')/RC}\left(\int\limits_{0}^{t}\int\limits_{0}^{t'} \eta(\hat{t})\eta(\hat{t'})e^{(\hat{t}+\hat{t'})/RC}d\hat{t}d\hat{t'} + 2U_0\int\limits_{0}^{t} \eta(\hat{t})e^{\hat{t}/RC}d\hat{t} + U_0^2\right)\right\rangle
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=e^{-(t+t')/RC}\left(\int\limits_{0}^{t}\int\limits_{0}^{t'} \frac{2k_BT}{RC^2}\delta(\hat{t}-\hat{t}') e^{(\hat{t}+\hat{t'})/RC}d\hat{t}d\hat{t'} + U_0^2\right)$
To be correct, we have to treate two different cases, one for $t<t'$ and one for $t>t'$. But let's focus on $t>t'$ at the moment. Then the next step would be
$\langle U(t)U(t')\rangle =e^{-(t+t')/RC}\left(\frac{2k_BT}{RC^2}\int\limits_{0}^{t}e^{2\hat{t}/RC}d\hat{t} + U_0^2\right)
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=e^{-(t+t')/RC}\left(\frac{k_BT}{C}e^{2t/RC} + U_0^2\right)
%
=\frac{k_BT}{C}e^{(t-t')/RC} + U_0^2e^{-(t+t')/RC}$.
For $t<t'$ you'll get
$\langle U(t)U(t')\rangle =\frac{k_BT}{C}e^{(t'-t)/RC} + U_0^2e^{-(t+t')/RC}$.
So we can combine both results to
$\langle U(t)U(t')\rangle =\frac{k_BT}{C}e^{-|t-t'|/RC} + U_0^2e^{-(t+t')/RC}$.
If you choose the initial voltage $U_0 = 0$, this gives you the exakt result from wikipedia.
