Does the lognormal distribution satisfy the Fokker-Planck equation for geometric brownian motion?

Question

I've been trying to understand the Fokker-Planck equation and have hit a road block on what is basically the second simplest possible example, geometric Brownian motion.

Consider this Ito SDE for GBM, where $W$ is Brownian motion: $$ dX_t = \mu X_t dt + \sigma X_t dW_t $$ $$ X_0 = 1 $$

So the Fokker-Planck equation $$ 0 = \frac{\partial{p}}{\partial{t}} + \frac{\partial{}}{\partial{x}} \left[ \alpha(x,t) p(x,t) \right] - \frac{\partial{}^2}{\partial{x}^2} \left[ \frac{1}{2}\beta(x,t)^2 p(x,t) \right] $$ should hold for $\alpha(x,t) = \mu x $ and $\beta(x,t)=\sigma x $, and $p(x,t)$ equal to the lognormal density function with parameters $\mu t$ and $\sigma \sqrt{t}$

I've attempted to verify this by plugging it all into Mathematica:

fp[a_,b_,p_] := D[p[x,t],t] + D[a[x,t]*p[x,t],x]-D[(1/2)*b[x,t]^2*p[x,t],{x,2}]
FullSimplify[
  fp[
    Function[{x,t},\[Mu]*x ],
    Function[{x,t},\[Sigma]*x],
    Function[{x,t},PDF[LogNormalDistribution[ \[Mu]*t, \[Sigma]*Sqrt[t]]][x]]
  ],
  {x > 0, \[Sigma]>0,t>0}
]

This is the output I get: $$ \frac{ e^{\frac {(-t \mu + Log[x])^2}{2 t \sigma^2} }(-t \mu + Log[x]) }{ 2 \sqrt{2 \pi} t^{3/2} x \sigma } $$

I was expecting to get zero. Any ideas about where I could be going wrong here?