Let y = y(x) be the solution of the differential equation ${{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0$ then, the minimum value of $y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$ is equal to :