Let C1  be the curve obtained by the solution of the differential equation 2xydydx=y2−x2Let the curve C2 be the solution of 2xyx2−y2=dydx.If both curves pass through the point 1,1,then the area enclosed by the curve C1  and C2  is equal to

The given differential equation dydx=y2-x22xy, it is homogeneous equation.  Put y=vxv+xdvdx=v2x2-x22vx2=v2-12vxdvdx=v2-1-2v22v=-v2+12v⇒2vv2+1dv=-dxxlnv2+1=-ℓnx+lnc⇒v2+1=cx⇒y2x2+1=cx⇒x2+y2=cx If pass through (1,1)  ∴x2+y2-2x=0Similarly second differential equation is dxdy=x2−y22xyEquation of curve is x2+y2−2y=0 Area of the region required is equal to the double the ( area of the quarter circle - area of the triangle )=14×π×12−12×1×1×2=π2−1 square units