Solve xp²−2yp+ax=0

The equation xp² − 2yp + ax = 0 (p = dy/dx) is solvable for y.

Solve for y

2yp = xp² + ax → y = x(p² + a)/(2p) = x(p/2 + a/(2p))

Differentiate w.r.t. x

p = (p² + a)/(2p) + x(1/2 − a/(2p²)) dp/dx

p − (p² + a)/(2p) = (p² − a)/(2p), so

(p² − a)/(2p) = x (p² − a)/(2p²) · dp/dx

Cancel (p² − a)/(2p) (the case p² = a is treated below):

1 = (x/p) dp/dx → dp/p = dx/x → p = cx

Eliminate p

y = x(c²x² + a)/(2cx) = (c²x² + a)/(2c)

General solution: 2cy = c²x² + a

Check: p = cx, then xp² − 2yp + ax = c²x³ − (c²x² + a)x + ax = 0 ✓.

Singular solution

The factor p² = a gives y = ±√a x, which satisfies the given equation. It is the envelope of the family 2cy = c²x² + a: treating it as a quadratic in c, c²x² − 2cy + a = 0, the discriminant vanishes when 4y² − 4ax² = 0, i.e.

y² = ax² (singular solution)