If $X_t$ is an Ito process, such that:
$dX_t = \mu(t, X_t)dt + \sigma(t, Xt)dW_t$ where $W_t$ is a standard brownian motion.
Then we can say that:
$E(dX_t) = \mu(t, X_t)dt$ and that $Var(dX_t) = \sigma^2(t, Xt)dt$
Is this equivalent to saying that (I removed the differential operator):
$E(X_t) = \mu(t, X_t)\times t$ and that $Var(X_t) = \sigma^2(t, Xt)\times t$