JEE Main 20262 Apr, Evening shiftMathematicsEasyDifferential Equations163× since 2002Let x=x(y)x=x(y)x=x(y) be the solution of the differential equation 2y2dxdy−2xy+x2=02y^2\frac{dx}{dy}-2xy+x^2=02y2dydx−2xy+x2=0, y>1y>1y>1, x(e)=ex(e)=ex(e)=e. Then x(e2)x(e^2)x(e2) is equal to :AOption A. 32e2\frac{3}{2}e^223e2BOption B. 23e2\frac{2}{3}e^232e2COption C. e2e^2e2DOption D. 2e22e^22e2CheckSkip · show solution
