Singular solution of the differential equation $(y')^2-3xy'+y^2=0$.

Question

To find singular solution of the differential equation $$(y')^2-3xy'+y^2=0$$ I find its $p $-discriminant as $$y=\pm \frac {3}{2}x $$ but none of the factor satisfies differential equation. How to find its singular solution? Thank you .

Answer 1

Your equation has one clear singular solution which is $y(x) = 0$. It may also have other singular solutions, but if it does these are not so easy to find. You are correct to use the $p$-discriminant method, but in general solving via this method doesn't give singular solutions, only the locus of points in the curve $f(x,y,p) = 0$ which have constant $p$ (in your case, $f(x,y,p) := p^2-3xp+y^2$). So you shouldn't expect the $p-$discriminant that you found to solve the equation.

Based on my understanding from here, if you want to construct all singular solutions it's necessary to find both the $p$- and $c$-discriminants of your equation. This would require finding the general solution of your equation $\phi(x,y,c)$ in terms of integration constant $c$ to be able to find the $c$-discriminant. I spent a while trying to solve your equation and I couldn't find a general solution, and I also put it into Mathematica which doesn't know how to solve it either. I was surprised Mathematica didn't know, I guess the equation itself looks deceptively simple.

Here is my calculation of the $p-$discriminant, based on the analysis here. First $$\frac{\partial f}{\partial p} = 2p - 3x=0.$$ Then $$2 f - p \frac{\partial f}{\partial p} = 2 y^2 - 3 x p = 0\qquad\implies\qquad p = \frac{2 y^2}{3x}.$$Now substitute back into $f(x,y,p) = 0$ to find that $$\psi(x,y):=y^2 \left(\frac{4 y^2}{9 x^2}-1\right) = 0.$$ I believe that this $\psi(x,y)$ is the $p$-discriminant. I think importantly, this method avoids dividing by $y$ in coming to the solution, and we see that $y=0$ turns out to be a singular solution here, so we don't want to divide by $y$ while solving the system to arrive at the form for $\psi(x,y)$.

Then setting $\psi(x,y)$ to zero gives two solutions to the $p$-discriminant equation; the first one is $y(x) = 0$, and the second is $y(x)^2=\frac{9}{4}x^2$. This gives $|y(x)| = \pm \frac{3}{2} |x|$, which is almost what you have. But as I said, I think it's fine that this doesn't solve $f(x,y,y') = 0$.

Out of interest, where does this equation come from? Is it from some research problem, or from a course on differential equations?