Let \( y = y(x) \) be the solution of the differential equation \( \dfrac{dy}{dx} = 1 + x e^{y-x}, \\ -\sqrt{2} < x < \sqrt{2}, \\ y(0) = 0 \) then, the minimum value of \( y(x), x \in (-\sqrt{2 }, 2 )\) is equal to:
\( (1 + \sqrt{3}) - \log_e(\sqrt{3} - 1) \)
\( (2 + \sqrt{3}) + \log_e 2 \)
\( (1 - \sqrt{3}) - \log_e(\sqrt{3} - 1) \)
\( (2 - \sqrt{3}) - \log_e 2 \)