Solve x dy − y dx = √(x²+y²) dx

x dy − y dx = √(x² + y²) dx

The equation is homogeneous (take x > 0).

Substitution

Put y = vx, dy = v dx + x dv. Then x dy − y dx = x² dv, and the equation becomes

x² dv = x√(1 + v²) dx → dv/√(1 + v²) = dx/x

Integrate

sinh⁻¹ v = ln x + ln c → v + √(1 + v²) = cx

Replace v = y/x, so v + √(1 + v²) = (y + √(x² + y²))/x = cx, i.e.

y + √(x² + y²) = cx²

Alternative form

Isolating y: from √(x² + y²) = cx² − y, squaring gives x² = c²x⁴ − 2cx²y, so

2cy = c²x² − 1

(a family of parabolas).