MathematicsMedium38×since 2002Q3600If 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 :A2e\sqrt 2 e2eB123e{1 \over 2}\sqrt 3 e213eCe2{e \over {\sqrt 2 }}2eD3e\sqrt 3 e3eCheck answerSkip