Solve xy(p²+2)=2py³+x³

Note: the equation as printed, xy(p² + 2) = 2py³ + x³, does not factorise. The standard textbook equation it appears to come from is xy(p² + 2) = p(x² + 2y²), with p = dy/dx, whose right side was garbled in the scan. This evident intended version is solved here.

Factorise

xy p² + 2xy − p(x² + 2y²) = 0

xy p² − (x² + 2y²)p + 2xy = 0

(py − x)(px − 2y) = 0

(Check: expanding gives p²xy − 2py² − px² + 2xy ✓.)

Case 1: py − x = 0

y dy/dx = x → y dy = x dx → y² = x² + c

y² − x² = c

Case 2: px − 2y = 0

dy/y = 2 dx/x → ln y = 2 ln x + ln c

y = cx²

General solution

(y² − x² − c)(y − cx²) = 0, where c is an arbitrary constant (the same symbol is used for both factors, as is usual for such products).