JEE Main 201912 Apr, Morning shiftMathematicsMediumDifferential Equations163× since 2002Consider the differential equation, y2dx+(x−1y)dy=0{y^2}dx + \left( {x - {1 \over y}} \right)dy = 0y2dx+(x−y1)dy=0, If value of y is 1 when x = 1, then the value of x for which y = 2, is :AOption A. 32−1e{3 \over 2} - {1 \over {\sqrt e }}23−e1BOption B. 12+1e{1 \over 2} + {1 \over {\sqrt e }}21+e1COption C. 52+1e{5 \over 2} + {1 \over {\sqrt e }}25+e1DOption D. 32−e{3 \over 2} - \sqrt e23−eCheckSkip · show solution
