Expectation equation, harmonic functions, do not understand why equation is true 
Let $u: \mathbb{R}_+ \times \mathbb{R}^d$ be a bounded $C^2$ function whose first and second partial derivatives are uniformly bounded (or, more generally, have at most polynomial growth as $|x| \to \infty)$ on $[0, T] \times \mathbb{R}^d$, for any $0 \le T < \infty$. For any $t \ge 0$ and any $x \in \mathbb{R}^d$ it is not hard to see that,$$E^x u(t, W_t) = u(0, x) + E^x \int_0^y \left( {\partial\over{\partial s}} + {1\over2}\Delta_s\right) u(s, W_s)\,ds.$$

I don't see why this equality is true? Could anyone explain?
 A: This a simple consequence of Itô formula in stochastic calculus. I'll explain the one dimensional formula, the d-dimensional is similar.
First of all to answer the questions of Stefan and Wilie, $W_t$ is a Brownian motion process which takes the value $x$ at $t=0$ so the dependence of $x$ is implicit here, most of the references will consider $W_t$ a standard Brownian motion which means it takes the value $0$ at $t=0$ and in the above formula you'll have $x+W_t$ instead. $\Delta_s$ here means the sum of all the mixed derivatives.
Itô formula states that 
$$du(s,W_s)=\frac{\partial u}{\partial s}(s,W_s)ds+\frac{\partial u}{\partial x}(s,W_s)dW_s+\frac 12\frac{\partial^2 u}{\partial s^2}(s,W_s)ds  $$
Integrating this equation between $0$ and $t$ gives 
$$u(t,W_t)=u(0,W_0)+\int_0^t\frac{\partial u}{\partial s}(s,W_s)ds+\int_0^t\frac{\partial u}{\partial x}(s,W_s)dW_s+\int_0^t\frac12\frac{\partial^2 u}{\partial s^2}(s,W_s)ds $$
now taking the expactation $E^x$ which means the conditional expactation knowing that $W_0=x$ we get the results as we have 
$$ E^x[\int_0^t\frac{\partial u}{\partial x}(s,W_s)dW_s]=0$$
by the way, you have a typo in your formula where you wrote $y$ instead of $t$.
All the conditions about the polynomial growth are made just to make the expectations converge.
Hope this is clear.
