Ornstein-Uhlenbeck process steady state probability I need to find the steady state probability of an Ornstein-Uhlenbeck process. I've made a start, but I've got stuck at this point. First I start with the definition of the evolution of probability for the one variable Fokker-Planck equation:
$$\frac{\partial P}{\partial t}(x,t)= L_{FP} P(x,t)\\
L_{FP}=-\frac{\partial}{\partial x} D^{(1)}(x) + \frac{\partial^2}{\partial x^2} D^{(2)}(x)$$
where $P$ is the probability, and $D^{(1)}(x)$ and $D^{(2)}(x)$ are the drift and diffusion, respectively.
For an OU process,
$$D^{(1)}(x)=-\gamma x, \ D^{(2)}(x)=D = \text{const},$$
where $\gamma$ is a constant. Substituting this into the Fokker-Planck equation gives
$$\frac{\partial P}{\partial t}= \gamma\frac{\partial}{\partial x} (xP) + D\frac{\partial^2}{\partial x^2} P.$$
For a steady state solution, $\dfrac{\partial P}{\partial t}=0$, so the equation reduces to
$$\frac{\partial^2 P}{\partial x^2}+\frac{\gamma}{D}\frac{\partial}{\partial x} (xP) =0.$$
I'm guessing that now I'd expand the second term to give
$$\frac{\partial^2 P}{\partial x^2}+\frac{\gamma}{D}\left(x\frac{\partial P}{\partial x} + P\right)=0,$$
but then I have no idea where to go from there. I'm also guessing there will be some initial condition like
$$P(x,0) = \delta (x-x_0),$$
but I'm not sure. Can anyone help me please?
 A: I found the answer in the end. Thought I should post it here in case anyone runs into the same difficulties.
Starting from the above steady state, where technically the derivatives become normal derivates
$$ \frac{d^2 P}{d x^2}+\frac{\gamma}{D}\frac{d}{d x} (xP) =0 $$
... I rearrange then integrate
\begin{align}
\frac{d}{d x} (xP) =& -\frac{D}{\gamma}\frac{d^2 P}{d x^2}\\
xP =& -\frac{D}{\gamma}\frac{d P}{d x} + C_1.
\end{align}
Constant $C_1$ must be $0$ due to boundary conditions of $p,\frac{d p}{d x} \rightarrow 0$ as $x\rightarrow 0$.
Rearrange, integrate and rearrange:
$$ \int\frac{1}{P}\ dP = -\frac{\gamma}{D}\int x\  dx$$
$$ \ln P = -\frac{\gamma x^2}{2D} + C_2$$
$$ P = C_3 e^{-\frac{\gamma x^2}{2D}}.$$
Constant $C_3$ is found through the condition that $\int^{+\infty}_{-\infty}P(x)\ dx = 1$.
$$ C_3 \int^{+\infty}_{-\infty}e^{-\frac{\gamma x^2}{2D}}\ dx = 1 .$$
With a bit of change of variables $y^2=\frac{\gamma x^2}{2D}$
$$ C_3 \sqrt{\frac{2D}{\gamma}} \int^{+\infty}_{-\infty}e^{-y^2}\ dy = 1 .$$
The integral can be shown (Example: How do you integrate $e^{x^2}$?) to equal $\sqrt{\pi}$. So
$$ C_3  = \sqrt{\frac{\gamma}{2\pi D}}$$
and finally 
$$ P = \sqrt{\frac{\gamma}{2\pi D}} e^{-\frac{\gamma x^2}{2D}}.$$
