JEE Main 20245 Apr, Evening shiftMathematicsMediumDifferential Equations163× since 2002The differential equation of the family of circles passing through the origin and having centre at the line y=xy=xy=x is :AOption A. (x2−y2+2xy)dx=(x2−y2+2xy)dy\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2+2 x y\right) \mathrm{d} y(x2−y2+2xy)dx=(x2−y2+2xy)dyBOption B. (x2+y2−2xy)dx=(x2+y2+2xy)dy\left(x^2+y^2-2 x y\right) \mathrm{d} x=\left(x^2+y^2+2 x y\right) \mathrm{d} y(x2+y2−2xy)dx=(x2+y2+2xy)dyCOption C. (x2+y2+2xy)dx=(x2+y2−2xy)dy\left(x^2+y^2+2 x y\right) \mathrm{d} x=\left(x^2+y^2-2 x y\right) \mathrm{d} y(x2+y2+2xy)dx=(x2+y2−2xy)dyDOption D. (x2−y2+2xy)dx=(x2−y2−2xy)dy\left(x^2-y^2+2 x y\right) \mathrm{d} x=\left(x^2-y^2-2 x y\right) \mathrm{d} y(x2−y2+2xy)dx=(x2−y2−2xy)dyCheckSkip · show solution
