<s>[INST] 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?
 [/INST] From @Rebellos work $$\int \frac{dy}{y}=\int \frac{dp}{p}+\int \frac{dp}{p(2-ap^2)}$$
In the third integral let $p=1/t \implies -dt/t^2$, then $2t^2-a=u$, we get
$$\ln y= \ln p-\frac{1}{4}\ln u \implies y(p)=\frac{Cp\sqrt{p}}{(2-ap^2)^{1/4}}~~~~(1)$$
Inserting this in the given ODE, we get
$$x(p)=\frac{y+ayp^2}{3p}~~~~(2)$$
Eqs.(1) and (2) define parametric equation of  family of curves when $p$ acts like a parametre, $C$ is the arbitrary constant of integration for a fixed value of $a$.
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