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 :

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 :

$\left( {2 - \sqrt 3 } \right) - {\log _e}2$

$\left( {2 + \sqrt 3 } \right) + {\log _e}2$

$\left( {1 + \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$

$\left( {1 - \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)$

Solution

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