= Solution
The pair $(v_t,Y_t)$ is a <Markov diffusion> with infinitesimal generator
$$
\mathcal L
=(\alpha-\beta v)\partial_v
+\frac12\gamma^2v\,\partial_{vv}
+v\,\partial_y.
$$
Part d shows that the terminal condition in the equation for $\widetilde U$ is exactly
$$
G\bigl(\widetilde S_T,(1-\rho^2)Y_T\bigr)
$$
when evaluated at $(v_T,Y_T)$. The <Feynman-Kac formula> applied to the displayed backward equation therefore gives
$$
\widetilde U(0,v_0,0)
=\mathbb E\!\left[
G\bigl(\widetilde S_T,(1-\rho^2)Y_T\bigr)\right].
$$
Part c identifies the right side with $\pi_0$.
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