Singular solution of first order differential equation

Question

The singular solution of $y = px + p^3,$ where, $ p = dy/dx$. I have the answer

$4x^3 + 27y^2 = 0.$ as given in the key, but I am not getting this answer when using $p$- discriminant method.

Answer 1

$$y = px + p^3\tag1$$which is of Clairaut's form and hence the general solution is $$y=cx+c^3\qquad\text{, where $~c~$ is an arbitrary constant.}\tag2$$ Since the given equation is of Clairaut's form, $p-$discriminant and $c-$discriminant will be exactly same. Hence we will find the only $p-$discriminant.

Differentiating $(1)$ by $~p~$, we have $$x+3p^2=0~.\tag3$$ which implies that the domain of the singular solution is $x≤0~$, there should be no complex values in a real problem.

The differential equation $(1)$ can be written as, $$y=px+p^3=\frac13p(x+3p^2)+\frac23px=\frac23px\qquad\text{(using $(3)$)}$$ Squaring both side, $$y^2=\frac49p^2x^2= \frac49~x^2~\left(-\dfrac x3\right)=-\frac4{27}~x^3\qquad\text{(using $(3)$)}$$ $$\implies 4x^3+27y^2=0$$ which is the required singular solution.

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thanks to Lutz Lehmann.