JEE Math Practice Question

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 :

1. A.$$\left( {2 - \sqrt 3 } \right) - {\log _e}2$$
2. B.$$\left( {2 + \sqrt 3 } \right) + {\log _e}2$$
3. C.$$\left( {1 + \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$$
4. D.$$\left( {1 - \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$$