Stationary distribution for Kolmogorov Forward Equation

Question

Given $X_t$ which satisfies the following SDE,

$$ dX_t = f'(X_t)dt + \sigma dW_t $$

where f is an infinitely differentiable function, and $f'$ above is the first derivative of $f$.

We know that the Kolmogorov Forward Equation is,

$$ \frac{\partial P}{\partial t} + \frac{\partial{(pf')}}{\partial x} - \frac{1}{2}\sigma^2\frac{\partial^2P}{\partial^2 x} = 0 $$

How do i show that the $Ae^{2f(x)/\sigma^2}$ is a solution for the stationary distribution?

What is did was the following,

The stationary distribution is given by the solution to,

$$ \frac{\partial{(Pf')}}{\partial x} - \frac{1}{2}\sigma^2\frac{\partial^2P}{\partial^2 x} = 0 $$

So we substitute $Ae^{2f(x)/\sigma^2}$ into the above equation to get,

$$ \begin{align} &f''(x)Ae^{2f(x)/\sigma^2} + 2f'(x)^2/\sigma^2 Ae^{2f(x)/\sigma^2}-\frac{1}{2}\sigma^2\left( 4f'(x)^2/\sigma^4 \right)Ae^{2f(x)/\sigma^2}\\ &=f''(x)Ae^{2f(x)/\sigma^2} \end{align} $$

which doesn't seem to solve the equation correctly.

Did I make some silly mistake somewhere?

Answer 1

$$\frac{d P}{d x} = \frac{2f'(x)}{\sigma^2}P$$ $$\frac{d^2 P}{d x^2} = \frac{2f''(x)}{\sigma^2}P + \frac{2f'(x)}{\sigma^2}P' = \frac{2f''(x)}{\sigma^2}P + \frac{4(f'(x))^2}{\sigma^4}P$$

It appears that you included the second term in the above derivation but forgot the first term. Adding it in gives you zero.