JEE Main 202224 Jun, Morning shiftMathematicsMediumDifferential Equations163× since 2002If x = x(y) is the solution of the differential equation ydxdy=2x+y3(y+1)ey, x(1)=0y{{dx} \over {dy}} = 2x + {y^3}(y + 1){e^y},\,x(1) = 0ydydx=2x+y3(y+1)ey,x(1)=0; then x(e) is equal to :AOption A. e3(ee−1){e^3}({e^e} - 1)e3(ee−1)BOption B. ee(e3−1){e^e}({e^3} - 1)ee(e3−1)COption C. e2(ee+1){e^2}({e^e} + 1)e2(ee+1)DOption D. ee(e2−1){e^e}({e^2} - 1)ee(e2−1)CheckSkip · show solution
