I am trying to find the general solution, $p$ and $c$ discriminant and hence the singular solutions (if any) of the following D.E. where $p=\mathrm dy/\mathrm dx$. $$xp^2-2py+4x=0$$
The obvious thing to do is isolate $p$ using the quadratic formula. The $p$-discriminant is $4y^2-16x^2$ so I suspect one or both of $y=\pm 2x$ could be the envelope of the general solution (i.e., singular soln. of our D.E.). $$p=\frac{y}{x}\pm \sqrt{\left(\frac yx\right)^2-4}$$
Substitute $y=tx$ to get: $$x\cdot\frac{\mathrm dt}{\mathrm dx}=\pm\sqrt{t^2-4}$$ Integrating: $$\ln|cx|=\pm \ln\left|t+\sqrt{t^2-4}\right|$$ where $c$ is the parameter of the general soln. Now we have two cases:
- $$cx=t+\sqrt{t^2-4}\\ \implies cx^2=y+\sqrt{y^2-4x^2}$$
- $$cx=\frac{1}{t+\sqrt{t^2-4}}=\frac{t-\sqrt{t^2-4}}4\\ \implies cx^2=y-\sqrt{y^2-4x^2}\\ \text{where } 4c\mapsto c$$
I am unable to decide which one to go with for finding the $c$-discriminant. Is there any other way to solve this D.E. instead of isolating $p$ by quadratic formula?
Update: I realized I had made a lot of silly mistakes. I've fixed them. Now it's clear that collectively, the two cases yield (take $c/2\mapsto c$): $$c=\frac{y\pm\sqrt{y^2-4x^2}}{2x^2}\\ =\frac{-(-y)\pm\sqrt{(-y)^2-4x^2\cdot 1}}{2x^2}$$ Using the quadratic formula (in reverse), the $c$-discriminant turns out to be $y^2-4x^2$. This confirms that $y=\pm 2x$ are indeed envelops and hence, singular solutions. $$\phi(x,y,c)=x^2c^2-yc+1=0$$ See this Desmos slideshow (graph of the envelope and the family $\phi$).
Alternate approach: Isolate $y$ in the original D.E. $$y=\frac{px}{2}+\frac{2x}{p}$$ This seems a little off from the Clairut's form i.e., $y=px+f(p)$. Substitute $u=x^2$. Denote $\mathrm dy/\mathrm du$ by $m$. So differentiating w.r.t. $y$, we have $\frac{1}{m}=\frac{2x}p$ i.e., $p=2mx$. $$y=um+\frac{1}{m}$$ This is in Clairut's form i.e., $y=um+f(m)$ so the general soln. is given by $y=uc+f(c)$ i.e., $$y=uc+\frac{1}{c}=cx^2+\frac{1}{c}$$ And the singular soln. can be determined by eliminating $m$ from the D.E. using $u+f'(m)=0$ i.e., $m=\pm1/\sqrt u$. $$y=\pm 2\sqrt u=\pm 2x$$
Try polynomials of degree $2$ to get the general solution $$ y = c x^2 + \frac{1}{c}$$ and singular solutions $y = \pm 2 x$.