JEE Main 20209 Jan, Evening shiftMathematicsMediumDifferential Equations163× since 2002If dydx=xyx2+y2{{dy} \over {dx}} = {{xy} \over {{x^2} + {y^2}}}dxdy=x2+y2xy; y(1) = 1; then a value of x satisfying y(x) = e is :AOption A. 2e\sqrt 2 e2eBOption B. 123e{1 \over 2}\sqrt 3 e213eCOption C. e2{e \over {\sqrt 2 }}2eDOption D. 3e\sqrt 3 e3eCheckSkip · show solution
