How to solve this differential equation? (Steady State Solution of Forward Kolmogorov Equation) Here's the full question and my attempt at answering it by solving the differential equation.
Consider the following SDE
$$
d\sigma = a(\sigma,t)dt + b(\sigma,t)dW
$$
The Forward Equation (FKE) is given by
$$
\frac{\partial p}{\partial t} = \frac{1}{2}\frac{\partial^2}{\partial\sigma^2}(b^2p) - \frac{\partial}{\partial\sigma}(ap)
$$
The steady state solution is given by setting $\frac{\partial p}{\partial t}=0$. Considering the boundary conditions that as $\sigma\rightarrow\infty$, $p\rightarrow 0$ and $\frac{\partial p}{\partial \sigma}\rightarrow 0$, show that the steady state solution is given by
$$
p(\sigma) = \frac{A}{b^2}e^{\int^\sigma\frac{2a}{b^2}d\sigma '}
$$
Essentially, this is a differential equation problem. Thing is, I'm stuck on how to solve it.
This is what I've got so far:
So, as per the question, the steady state solution is:
$$
\frac{1}{2}\frac{d^2}{d\sigma^2}(b^2p) - \frac{d}{d\sigma}(ap) = 0
$$
Integrating to convert to a first order differential equation?:
$$
\frac{1}{2}\frac{d}{d\sigma}(b^2p) - ap = A
$$
Given the boundary conditions in the question, $A = 0$??
$$
\frac{1}{2}\frac{d}{d\sigma}(b^2p) = ap
$$
Re-arranging we have a variable separable D.E
$$
\frac{1}{p} dp = \frac{2a}{b^2} d\sigma
$$
Integrating:
$$
\log p = \int \frac{2a}{b^2} d\sigma + D
$$
$$
p = e^{\int \frac{2a}{b^2} d\sigma + D}
$$
....and this is where I'm stuck. It looks a little like the expected solution but I don't know where the extra $\frac{A}{b^2}$ coefficient comes from and why there is a $d\sigma '$ in the expected solution.
Where have I gone wrong?
 A: From $$\frac{1}{2} \frac{d}{d\sigma} (b^2 \cdot p) = a \cdot p$$ you concluded $$\frac{1}{p} \, dp = \frac{2a}{b^2} \, d\sigma$$
But this is not correct: Note that $b$ does depend on $\sigma$. You have to apply the chain rule:
$$\begin{align} \frac{1}{2} \frac{d}{d\sigma} (b^2 \cdot p) &= a \cdot p \\ \Rightarrow b \cdot \frac{d b}{d \sigma} \cdot p + \frac{1}{2} b^2 \cdot \frac{dp}{d \sigma} &= a \cdot p \\ \Rightarrow \frac{1}{p} \, dp &= 2 \frac{ \left(a- b \cdot \frac{db}{d\sigma}\right) }{b^2} \, d\sigma  = \frac{2a}{b^2} \, d\sigma - 2 \frac{1}{b} \cdot \frac{db}{d\sigma} \, d\sigma \end{align}$$
From the second term on the righthand-side you obtain the factor $\frac{1}{b^2}$ in the given solution.
Concerning $d\sigma'$: Since $p=p(\sigma)$ you shouldn't use $\sigma$ as variable of integration at the same time. That's why they called it $\sigma'$ instead (has nothing to do with the derivative!).
A: Note that the boundary conditions
$$\sigma\rightarrow\infty\Rightarrow\left\{\begin{array}{c}p\rightarrow 0\\ \frac{\partial p}{\partial\sigma}\rightarrow 0\end{array}\right.$$
Are not enough to state
$$\frac{\partial}{\partial\sigma}(b^2p)-(ap)=A=0$$
Because, as $b$ is a function of $\sigma$,
$$\frac{\partial}{\partial\sigma}(b^2p) = 2b\frac{\partial b}{\partial\sigma}p+b^2\frac{\partial p}{\partial\sigma}$$
And you will also need that, as $\sigma\rightarrow \infty$, that
$$\begin{array}{c}bp\frac{\partial b}{\partial\sigma}\rightarrow 0\\b^2\frac{\partial p}{\partial\sigma}\rightarrow 0 \end{array}$$
as well to get $A=0$.
