Consider the nonlinear SDE for $(X_t)_{t\geq 0}$ \begin{equation} \mathop{dX_t}=X_t\left(\mu\mathop{dt}+\sqrt{v_0+\omega\left(\phi(t)-\ln X_t\right)^2}\mathop{dW_t}\right), \end{equation} where $\phi:t\rightarrow \mathbb{R}_{\geq 0}$ is a deterministic bijection and $\mu,v_0,\omega\in\mathbb{R}_{>0}$ are constants. Applying the variance (or Lamperti) transform \begin{align} f(x,t)&=\int\frac{1}{x\sqrt{v_0+\omega\left(\phi(t)-\ln x\right)^2}}\mathop{dx},\\ &=-\frac{1}{\sqrt{\omega}}\tanh^{-1}\left(\frac{\left(\phi(t)-\ln x\right)\sqrt{\omega}}{\sqrt{v_0+\omega\left(\phi(t)-\ln x\right)^2}}\right), \end{align} and Itô's Lemma should result in a linear SDE for $Y_t:=f(x,t)$ with variance $1$: \begin{align} \mathop{dY_t}=\frac{\partial f(X_t,t)}{\partial t}\mathop{dt}+\frac{\partial f(X_t,t)}{\partial X}\mathop{dX_t}+\frac{1}{2}\frac{\partial^2f(X_t,t)}{\partial X^2}\left(\mathop{dX_t}\right)^2. \end{align} Is it possible to then derive the probability density function of $X_t$? Any help would be much appreciated.
Note that, using the Fokker-Planck equation, this is also equivalent to solving the PDE \begin{equation} p_t(x,t)=\frac{x^2p_x(x,t)(v_0+\omega(\phi(t)-\ln x)^2)}{2}+xp(x,t)\left(v_0+\omega\left((1-2\phi(t))\ln x+\ln^2 x +\phi(t)(1-\phi(t))\right)\right)-\mu xp_{x}(x,t)-\mu p(x,t), \end{equation} with initial condition $p(x,0)=\delta(x-X_0)$. However, this seems intractable and so I was wondering if applying an appropriate transformation would simplify the problem.