Let $h$ and $V$ be bounded continuous functions on $\mathbb{R}^d$. Suppose $u$ is continuous on $\mathbb{R}_+ \times \mathbb{R}^d$, bounded on $[0,T] \times \mathbb{R}^d$ for eacht $T < \infty$ and $u \in C^{1,2}((0,\infty) \times \mathbb{R}^d)$. Let $u$ satisfy the initial value problem \begin{align} \frac{\partial}{\partial t} u(t,x) &= \frac{1}{2} \Delta u(t,x) + V(x)u(t,x) \qquad \text{on $(0,\infty)\times \mathbb{R}^d$}\\ u(0,x) &= h(x), \qquad x \in \mathbb{R}^d, \end{align} where $\Delta u = u_{x_1,x_1} + \ldots + u_{x_d,x_d}$.
I want to show that \begin{align} u(t,x) = \mathbb{E}^x[h(B(t)) e^{\int_0^t V(B(s)) ds}], \end{align} where $B(\cdot)$ is Brownian motion on $\mathbb{R}^d$ and $\mathbb{E}^x$ is expectation under the path measure of $B(\cdot)$ started at $B(0)=x$.
Therefore, we consider the process \begin{align} Z_t = u(t_o-t, B(t)) e^{\int_0^t V(B(s)) ds} \qquad \text{for $t \in [0, t_0 - \epsilon]$, $\epsilon >0$.} \end{align} Now, I do not know how to apply Itô's Lemma to the process $Z_t$. What expression do we find according to Itô's Lemma for $dZ_t$?
According to the comment, we find that \begin{align} Z_t &= X_t Y_t \\ &= X_0Y_0 + \int_0^t X_s Y_s + \int_0^t Y_s X_s + [X,Y]_t. \end{align} Where $X_t = u(t_0-t,B_t)$ and $Y_t = e^{\int_0^t V(B(s)) ds}$, so $X_0 = u(t_0,B_0)= u(t_0,x) =^? h(x)$ and $Y_0 =1$.
The question remains, how do we find an expression for $dZ_t$?
