JEE Main 202125 Jul, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation dydx=1+xey−x,−2<x<2,y(0)=0{{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2 < x < \sqrt 2 ,y(0) = 0dxdy=1+xey−x,−2<x<2,y(0)=0 then, the minimum value of y(x),x∈(−2,2)y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)y(x),x∈(−2,2) is equal to :AOption A. (2−3)−loge2\left( {2 - \sqrt 3 } \right) - {\log _e}2(2−3)−loge2BOption B. (2+3)+loge2\left( {2 + \sqrt 3 } \right) + {\log _e}2(2+3)+loge2COption C. (1+3)−loge(3−1)\left( {1 + \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)(1+3)−loge(3−1)DOption D. (1−3)−loge(3−1)\left( {1 - \sqrt 3 } \right) - {\log _e}\left( {\sqrt 3 - 1} \right)(1−3)−loge(3−1)CheckSkip · show solution
