JEE Main 20205 Sep, Morning shiftMathematicsMediumDifferential Equations163× since 2002If y = y(x) is the solution of the differential equation 5+ex2+y.dydx+ex=0{{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 02+y5+ex.dxdy+ex=0 satisfying y(0) = 1, then a value of y(log_e13) is :AOption A. -1BOption B. 1COption C. 0DOption D. 2CheckSkip · show solution