Solve the equation $p^3x-p^2y-1=0$ where $p=\frac {dy}{dx}$
My Attempt:
$$p^3x-p^2y-1=0$$
$$p^2y=p^3x-1$$
$$y=px-\frac {1}{p^2}$$
This is solvable for $y$ so differentiating both sides w.r.t $x$
$$\frac {dy}{dx} = p + x \cdot \frac {dp}{dx} – (-2) p^{-3} \cdot \frac {dp}{dx}$$
$$p=p+x\cdot \frac {dp}{dx} + \frac {2}{p^3} \cdot \frac {dp}{dx}$$
$$x\cdot \frac {dp}{dx} + \frac {2}{p^3} \cdot \frac {dp}{dx}=0$$
$$(x+\frac {2}{p^3}) \cdot \frac {dp}{dx}=0$$
How do I solve further?
Best Answer
Note that $y=xp-\frac1{p^2}$ is already an instance of Clairaut's equation. As such, it has the solutions $$y=kx-\frac1{k^2}\qquad(x,y)=(-2/t^3,-3/t^2)$$ where the singular solution may be made explicit as $$y=-3\left(\frac x2\right)^{2/3}$$