JEE Main 20211 Sep, Evening shiftMathematicsMediumDifferential Equations163× since 2002If y = y(x) is the solution curve of the differential equation x2dy+(y−1x)dx=0{x^2}dy + \left( {y - {1 \over x}} \right)dx = 0x2dy+(y−x1)dx=0 ; x > 0 and y(1) = 1, then y(12)y\left( {{1 \over 2}} \right)y(21) is equal to :AOption A. 32−1e{3 \over 2} - {1 \over {\sqrt e }}23−e1BOption B. 3+1e3 + {1 \over {\sqrt e }}3+e1COption C. 3 + eDOption D. 3 −-− eCheckSkip · show solution
