I have a question that's driving me crazy.
I have a Fokker-Planck equation
$$\frac{\partial P}{\partial t}=x\frac{\partial P}{\partial x}+D\frac{\partial^2 P}{\partial x^2}$$
I look for the stationary solution
$$ x\frac{d P}{d x}+D\frac{d^2 P}{d x^2}=0$$
Now, I know that the solution is
$$P(x)=Ce^{-x^2/2D} $$
But I did not understand how to arrive to the solution, because my strategy would be to do a change of variable:
$$xY+D\frac{dY}{d x}=0$$
$$\frac{dP}{d x}=Y$$
So I obtain
$$Y=Y_0 e^{-x^2/2D}$$
$$P=Y_0\int e^{-x^2/2D}dx$$
I am sure that there is something that I miss somewhere, but I do not know where. Thank you for the help !
EDIT:
To be sure that I understand the multidimensional case: $$dA=f_1(A,B,C)dt+\sigma_1dW_1$$ $$dB=f_2(A,B,C)dt+\sigma_2dW_2$$ Is the FPE for $P(A,B,t)$ $$\frac{\partial P}{\partial t}=-\frac{\partial f_1 P}{\partial A}-\frac{\partial f_2 P}{\partial B}+\frac{1}{2}\sigma_1^2\frac{\partial^2 P}{\partial A^2}+\frac{1}{2}\sigma_2^2\frac{\partial^2 P}{\partial B^2}$$
Thanks a lot !!!
The equation you are solving is not the equation that you want to solve. The FPE for the one dimensional autonomous Ito SDE
$$dX=b(X) dt + \sigma(X) dW$$
is
$$\frac{\partial P}{\partial t}=-\frac{\partial}{\partial x}(b(x) P(x,t))+\frac{\partial^2}{\partial x^2} \left ( \frac{\sigma^2}{2} P(x,t) \right ).$$
In your case, taking $D=1$, $b(x)=-x,\sigma(x)=\sqrt{2}$, so the stationary FPE reads
$$\frac{\partial}{\partial x}(x P(x))+\frac{\partial^2}{\partial x^2}(P(x))=0$$
which does hold with $P(x)=Ce^{-x^2/2}$. You can solve this basically doing what you did in reverse: integrate the derivatives away and then solve a first order linear equation with a constant forcing (but where the constant is unknown). Then use the fact that $P$ must be a probability density to identify the constants of integration.