Suppose $\delta$ is a vector in $\mathbb{R}^d$ with all the components being positive. Assume that $x\in\mathbb{R}^d_+$ and $Q(x)$ denotes the diagonal matrix formed by the vector $x$. It is known that the solution of the following PDE $$\frac{\partial u(t,x)}{\partial t}=\frac{1}{2}\operatorname{Trace}(Q(x)D_x^2u(t,x))+\langle\delta,D_x u(t,x)\rangle \\ u(0,x)=f(x)$$ is given by the following function: $$\int_{\mathbb{R}^d_+} f(y)\frac{1}{(2t)^d}\prod_{i=1}^d\left[\left(\frac{x_i}{y_i}\right)^{-\nu_i}e^{-\frac{x_i+y_i}{2t}}I_{\nu_i}\left(\frac{\sqrt{x_iy_i}}{t}\right)\right]\,dy \ ,$$ where $\nu_i=\frac{\delta_i}{2}-1$ and $I_\mu$ denotes the modified Bessel function with parameter $\mu$. Now, consider the following PDE: $$\frac{\partial v(t,x)}{\partial t}=\frac{1}{2}\operatorname{Trace}(e^{tB}Q(e^{-tB}x)e^{tB^*}D^2v(t,x))+\langle e^{tB}\delta, D_xv(t,x) \rangle \\ v(0,x)=f(x) \ \ , $$ where $B$ is a matrix and $Q(e^{-tB}x)$ is the diagonal matrix formed by the vector $e^{-tB}x$.
Question: Is it possible to somehow relate the solution of the second PDE to that of the first one (which is written in term of the integral)? Any reference about this literature would also be helpful.