Differential equation for Brownian motion I was trying to solve the following differential equation:
$$ \frac{dv}{dt} + \frac{\gamma}{m}v + \omega_0^2x = \frac{1}{m} \psi(t)$$
where: $v=\frac{dx}{dt}$ and $x=x(t)$. But I cant really advance.
The solution for it looks like this:
$$ v(t)=v_0 e^{- \Gamma t}C(t) - \frac{\omega_0^2}{\Delta}x_0e^{- \Gamma t} \sinh(\Delta t)+ \frac{1}{m} \int_0^tdt' \psi(t')e^{- \Gamma (t-t')}C(t-t') \hspace{1cm}(1)$$
where: $C(t)= \cosh(\Delta t )- \frac{\Gamma}{\Delta}\sinh(\Delta t)$ , $\Gamma = \frac{\gamma}{m}$ and $\Delta = \sqrt{\Gamma^2 - \omega_0^2}$.
I tried the annihilation polynomial (by, first, assuming the homogeneous equation):
$$ P(\lambda) = \lambda^2+ \frac{\gamma}{m} \lambda + \omega_0^2 =0$$
which lead me to:
$$ \lambda_{\pm} = - \frac{\Gamma}{2} \pm \sqrt{ \frac{\Gamma^2}{4}- \omega_0^2  }$$
So, the homogeneous solution would be:
$$ x(t) = c_1e^{ - \frac{ \Gamma}{2}t} \cos \left(\sqrt{ \frac{\Gamma^2}{4}- \omega_0^2  } t \right) + c_2e^{ - \frac{ \Gamma}{2}t} \sin \left(\sqrt{ \frac{\Gamma^2}{4}- \omega_0^2  } \right)$$
I dont know where the integration comes from, in the last term of $(1)$ on the RHS, since I would have to apply a derivative to this solution I obtained, in order to get the expression for the velocity $v(t)$.
I still need to find the particular solution, and I dont know how to find it either.
How can I obtain $(1)$?
 A: For the ODE
\begin{align}\tag{1}\label{1}
x''+\Gamma x'+\omega_0^2x=\frac{1}{m}\psi(t)
\end{align}
there are many ways of going about finding the solution. One method is using the particular solution and homogeneous solution to construct the general solution $y=y_h+y_0$, as I believe you are hoping to do. This of course only works if you know a particular solution already, which for arbitrary $\psi(t)$ you do not.
Here's one method that works for equations of the form $y''+by'+cy=f(x)$, where $b$ and $c$ are constants and $f(x)$ is an arbitrary function.
Taking $z=\delta x+x'$, $z'=\delta x'+x''$, where
\begin{align}
\delta^2-\Gamma \delta+\omega_0^2=0,\quad \rightarrow\quad \delta =\frac{1}{2}\left(\Gamma\pm\sqrt{\Gamma^2-4\omega_0^2}\right)
\end{align}
and substituting for $x$ and $x''$ into \ref{1} gives the equation
\begin{align}
z'+\frac{\omega_0^2}{\delta}z+(\delta^2-\Gamma\delta+\omega_0^2)x'=\frac{1}{m}\psi(t),\\
\longrightarrow \quad z'+\frac{\omega_0^2}{\delta}z=\frac{1}{m}\psi(t).
\end{align}
This is a first order linear ODE. Using the integration factor $\exp(\omega_0^2t/\delta)$ we arrive at
\begin{align}
z&=e^{-\omega_0^2t/\delta}\left[z_0+\frac{1}{m}\int_0^t e^{\omega_0^2t'/\delta}\psi(t')\mathrm dt'\right],\\
x'+\delta x&=e^{-\omega_0^2t/\delta}\left[v_0+\delta x_0+\frac{1}{m}\int_0^te^{\omega_0^2t'/\delta}\psi(t')\mathrm dt'\right].
\end{align}
Using the integrating factor $\exp(\delta t)$,
\begin{align}
x&=e^{-\delta t}\left[x_0+\frac{\delta(v_0+ \delta x_0)}{\delta^2-\omega_0^2}e^{(\delta-\omega_0^2/\delta)t}+\frac{1}{m}\int_0^te^{(\delta-\omega_0^2/\delta)t'}\int_0^{t'}e^{\omega_0^2t''/\delta}\psi(t'')\mathrm dt''\mathrm dt'\right],\\\\
x&=x_0e^{-\delta t}+\frac{\delta(v_0+\delta x_0)}{\delta^2-\omega_0^2}e^{-\omega_0^2t/\delta}+\frac{e^{-\delta t}}{m}\int_0^te^{(\delta-\omega_0^2/\delta)t'}\int_0^{t'}e^{\omega_0^2t''/\delta}\psi(t'')\mathrm dt''\mathrm dt'.\tag{2}\label{2}
\end{align}
I've checked that this satisfies the ODE by plugging it back in (hopefully no errors). This isn't quite the same form that your answer is given in, and I've given up trying to show that they are equivalent. I can't convince myself that the solution provided to you is correct, seeing as how when you take a derivative of equation (\ref{2}) you are left with a double integral, as well as the provided solution stating that $\Delta=\sqrt{\Gamma^2-\omega_0^2}$, when it should be $\Delta=\sqrt{\Gamma^2-4\omega_0^2}$.
