I am trying to understand this proof so that I can do the exercises without having to actually memorise the formula and plug in numbers, like a lot of people do. Thanks a lot in advance!
So if we have the forward Kolmogorov equation $\partial_t p=\partial_x(bp)+\frac{1}{2}\sigma^2\partial_{xx}p$, we can get the stationary distribution by solving $$\partial_x(bp)+\frac{1}{2}\sigma^2\partial_{xx}p=0.$$ This leads to $bp+\frac{1}{2}\sigma^2\partial_{x}p=C_1=0$ when the vanishing boundary condition at $\infty$ is assumed.
This is what I don't understand.
Separating variables $$p=p_s(x)=C_2e^{-2\int\frac{b}{\sigma^2}dx},$$ where $C_2$ is a constant.
How did it get there though?
$$ bp+\frac{1}{2}\sigma^2\partial_{x}p = 0 $$ $$ 2bp+\sigma^2\partial_{x}p = 0 $$ $$ \sigma^2\partial_{x}p = -2bp $$ $$ \frac{1}{p}\,\partial_{x}p = -\frac{2b}{\sigma^2} $$ $$ \partial_{x}\left(\ln p\right) = -\frac{2b}{\sigma^2} $$ $$ \ln p(x) = \ln p(A) - \int_{A}^{x}\frac{2b}{\sigma^2}\,dx' $$ where $A$ is a constant $$ p = p(A)\,\exp\left(- \int_{A}^{x}\frac{2b}{\sigma^2}\,dx'\right) $$
Now let $C \equiv p(A)$.