The curve satisfying the differential equation (x²−y²)dx+2xydy=0 and passing through the point (1,1) is:

(x²−y²)dx+2xydy=0⇒dy/dx=(y²−x²)/2xy
Put y=vx
dy/dx=v+x(dv/dx)
⇒v+x(dv/dx)=(v²x²−x²)/2vx²
⇒v+x(dv/dx)=(v²−1)/2v
⇒x(dv/dx)=(v²−1)/2v−v
⇒x(dv/dx)=−(v²+1)/2v
⇒2vdv/(v²+1)=−dx/x
Integrating we get;
ln|v²+1|=−ln|x|+lnc
y²/x²+1=c
Putting (1,1) c=2
x²+y²=2x=0
hence its is a circle of radius 1