`dy/dx=2xy^2/y-x^2`

Solve: xy(dy)/(dx)=x^2-y^2

C_1 and C_2 are two curves intersecting at (1,1) , C_1 satisfy dy/dx=(y^2-x^2)/(2xy) and C_2 satisfy dy/dx=(2xy)/(-y^2+x^2) then area bounded by these two curves is

x^2(dy/dx)^2-2xy dy/dx+2y^2-x^2=0

dy/dx=(2xy)/(x^2-1-2y)

(dy)/(dx)=xy+2y+x+2

Solve -x^(2)(dy)/(dx)=2xy+y^(2)

(dy)/(dx)=(2xy)/(x^(2)-y^(2))

The function y=f(x) is the solution of the differential equation (dy)/(dx)=(2xy+y^(2))/(2x^(2)) If f(1)=2, then the value of f((1)/(4)) is

(a) dy/dx = (xy)/(x^2+y^2)

2xy(dy)/(dx)=x^(2)+y^(2)