= Solution
For a separated weighted density, write $P_n=e^{\gamma t}p(\sigma)$. Its <eigenvalue equation> is
$$
\frac\kappa2p''+\frac1\tau(\sigma p)'+n\sigma p=\gamma p.
$$
Set $p=\psi\exp[-\sigma^2/(2\kappa\tau)]$. Differentiation eliminates the first derivative of $\psi$, giving the <tilted Ornstein-Uhlenbeck oscillator transformation>
$$
\frac\kappa2\psi''+\left[\frac1{2\tau}-\frac{\sigma^2}{2\kappa\tau^2}+n\sigma\right]\psi=\gamma\psi.
$$
Complete the square and introduce the <dimensionless variable> $x=(\sigma-\kappa n\tau^2)/\sqrt{\kappa\tau}$. The equation becomes
$$
\frac{d^2\psi}{dx^2}+\left[1+\kappa n^2\tau^3-2\gamma\tau-x^2\right]\psi=0.
$$
Comparing with the <quantum harmonic oscillator> gives $2E=1+\kappa n^2\tau^3-2\gamma\tau$. Its energies $E_m=m+1/2$ therefore correspond to $\gamma_m=\kappa n^2\tau^2/2-m/\tau$. The dominant nonnegative separated mode is the <ground state>, which is <nodeless>. Consequently
$$
\boxed{\gamma_n=\frac12\kappa\tau^2n^2.}
$$
For an independent check, $\log[\widetilde B(t)/B_0]=\int_0^t\widetilde\sigma(s)\,ds$ is a centered <Gaussian random variable> with <variance>
$$
V(t)=\kappa\tau^2\left[t-2\tau(1-e^{-t/\tau})+\frac\tau2(1-e^{-2t/\tau})\right].
$$
The <exponential moment of a Gaussian linear functional> gives $\mathbb E[B^n]=B_0^n\exp[n^2V(t)/2]$, whose long-time growth rate is the same $\gamma_n$. This exact <Ornstein-Uhlenbeck multiplicative amplification> formula also distinguishes the finite-time transient from the asymptotic exponential law. A general transient can contain sign-changing excited eigenfunctions in its expansion while the total weighted density remains nonnegative.
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