Solve y−3px+ayp²=0

y − 3px + ayp² = 0, with p = dy/dx.

Note: unlike the similar equation y − 2px + ayp² = 0, this one does not give a closed form in x and y. It is solved as written, and the solution is parametric in p.

Solve for x

3px = y(1 + ap²) → x = y(1 + ap²)/(3p)

Differentiate w.r.t. y

dx/dy = 1/p:

1/p = (1 + ap²)/(3p) + y(−1/(3p²) + a/3) dp/dy

1/p − (1 + ap²)/(3p) = (2 − ap²)/(3p), and −1/(3p²) + a/3 = (ap² − 1)/(3p²). Hence

(2 − ap²)/(3p) = y (ap² − 1)/(3p²) · dp/dy

The variables separate:

dy/y = (ap² − 1)/[p(2 − ap²)] dp

Integrate

By partial fractions, (ap² − 1)/[p(2 − ap²)] = −1/(2p) + (a/2)·p/(2 − ap²). So

ln y = −½ ln p − ¼ ln(2 − ap²) + ln C

y⁴ p² (2 − ap²) = C₁

Solution (parametric in p)

From the integral, y = C₁^(1/4) p^(−1/2)(2 − ap²)^(−1/4), and x follows from x = y(1 + ap²)/(3p).

x = y(1 + ap²)/(3p), y⁴p²(2 − ap²) = C₁

Extra solutions

If 2 − ap² = 0 (p = ±√(2/a), constant), the separated equation is satisfied with dp = 0, and the original equation becomes 3(y − px) = 0. So y = ±√(2/a) x are also solutions (these are lines through the origin).