Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$ Solve the equation $y-3px+ayp^2=0$ where $p=\frac {dy}{dx}$
My Attempt:
Given
$$y-3px+ayp^2=0$$
$$3px=y+ayp^2$$
$$x=\frac {1}{3} \cdot \frac {y}{p} + \frac {a}{3} \cdot yp$$
This is solvable for x. Differentiating both sides with respect to $y$
$$\frac {dx}{dy}=\frac {1}{3} \cdot \frac {p-y\frac {dp}{dy}}{p^2} + \frac {a}{3} (y\cdot \frac {dp}{dy} +p)$$
$$\frac {dx}{dy} = \frac {1}{3p} - \frac {y}{3p^2} \cdot \frac {dp}{dy} + \frac {ay}{3} \cdot \frac {dp}{dy} + \frac {ap}{3}$$
$$\frac {1}{p} - \frac {1}{3p} -\frac {ap}{3} = \frac {y}{3} \cdot \frac {dp}{dy}\cdot \frac {ap^2-1}{p^2}$$
$$\frac {3-1-ap^2}{3p} = \frac {y}{3} \cdot \frac {ap^2-1}{p^2} \cdot \frac {dp}{dy}$$
$$(2-ap^2) = \frac {y}{p} \cdot (ap^2-1) \cdot \frac {dp}{dy}$$
$$\frac {dy}{y} = \frac {ap^2-1}{p(2-ap^2)} dp$$
How do I integrate the RHS of above equation?

Solving for $p$ we get
$$
y' = \frac{3x\pm\sqrt{9x^2-4ay^2}}{2ay}
$$
or
$$
2ay y'=3x\pm\sqrt{9x^2-4ay^2}
$$
now calling $u = a y^2$ we have
$$
u' = 3x\pm\sqrt{(3x)^2-4u}
$$
and now following with $z = (3x)^2-4u$ we arrive at
$$
z'\pm 4\sqrt{z}=6x
$$
