Solve (x²−y²)dx+2xy dy=0

(x² − y²)dx + 2xy dy = 0

The equation is homogeneous of degree 2.

Substitution

Put y = vx, dy = v dx + x dv:

(x² − v²x²)dx + 2x(vx)(v dx + x dv) = 0

x²[(1 − v²) + 2v²]dx + 2x³v dv = 0

(1 + v²)dx + 2xv dv = 0

Separate variables

2v dv/(1 + v²) = −dx/x

ln(1 + v²) = −ln x + ln c

(1 + v²)x = c

Replace v = y/x:

x(1 + y²/x²) = c → x² + y² = cx

x² + y² = cx

This is a family of circles through the origin with centres on the x-axis.