Find the singular solution of the differential equation $$4xp^2=(3x-1)^2,$$ where $p=\frac{dy}{dx}.$
As we know the singular solution, of a first order differential equation, is represented by the envelope of the family of curves, represented by the general solution of the differential equation. So, first I tried calcculating the general solution of the differential equation, by first solving for $p.$
Given equation is :$$4xp^2=(3x-1)^2,$$ where $p=\frac{dy}{dx}.$
Rearranging this, we find, $$p=\frac{(3x-1)}{2\sqrt x}\implies \frac{dy}{dx}=\frac{(3x-1)}{2\sqrt x}\implies y=x\sqrt x-\sqrt x+c\implies (y-c)^2-x(x-1)^2=0,$$ where $c$ is the arbitrary constant of integration.
So, if $\phi(x,y,c)$ is the general solution of the differential equation, then $\phi(x,y,c)=(y-c)^2-x(x-1)^2=0.$
We now proceed to calculate the c-discriminant which is precisely, the eliminant between :
i.$\frac{\partial \phi}{\partial c}=0,$
ii.$\phi(x,y,c)=0.$
So, we get, $$\frac{\partial \phi}{\partial c}=\frac{\partial ((y-c)^2-x(x-1)^2)}{\partial c}=2(c-y)=0\implies c-y=0$$
and,
$\phi(x,y,c)=(y-c)^2-x(x-1)^2=0.$ Now, we eliminate $c$ as follows:
$(\frac{\partial \phi}{\partial c})^2-\phi(x,y,c)=x(x-1)^2=0.$ So, the c-discriminant is, $x(x-1)^2=0.$
After the calculation of the c-discriminant, we proceed to calculate the p-discriminant of $F(x,y,p)=0,$ where $F$ is the function representating the differential equation in the question. So, $F(x,y,p)=4xp^2-(3x-1)^2=0\implies 4xp^2-9x^2-1+6x=0.$
The p-discriminant is precisely, the eliminant between :
i.$\frac{\partial F}{\partial p}=0,$
ii.$F(x,y,p)=0.$
So, we get, $$\frac{\partial F}{\partial p}=\frac{\partial 4xp^2-9x^2-1+6x}{\partial p}=8xp=0\implies 4xp^2=0$$
and,
$F(x,y,p)=4xp^2-9x^2-1+6x=0.$ Now, we eliminate $p$ as follows:
$(\frac{\partial F}{\partial p})\frac p2-F(x,y,p)=4xp^2-(4xp^2-9x^2-1+6x)=(3x-1)^2.$ So, the p-discriminant is, $(3x-1)^2=0.$
So the p-discriminant, is, $(3x-1)^2=0.$
The c-discriminant,( which I calculalated previously,) is $x(x-1)^2=0.$
We use the fact,
A singular solution of a fist order differential equation satisfies both the p-discriminant and c-discriminant relations. In other words, if $P(x,y)=0$ satisfies is the p-discriminant relation, then $P(x,y)=hC_lT^3,$ where $h$ is the singular solution of the differential equation, $C_l$ is the "Cuspidal" Locus and T is the Tac locus of the given differential equation say, $f(x,y,p)=0.$
If $C(x,y)$ is the c-discriminant relation, then $C=hC_l^3N^2,$ where $C_l$ is the cuspidal locus and $N$ is the nodal locus.
In our case, we have, $P(x,y)=(3x-1)^2$ and $C(x,y)=x(x-1)^2.$
This is the point where I am stuck. How to find the singular solution using this particular approach ?Any help regarding this will be greatly appreciated.
For reference, one might check out this pdf on singular solutions of a 1st order differential equation.
In this pdf, as well, for some unspecified reason, they solved this exact same problem in a different method from the one, I wrote in here. (This problem is solved in the page number $9.$) I have no idea, why this method wasn't applied just for this case only! While, if one observes, in all the other cases, this method was utilised to find the $p$ discriminant. Is there a reason?