Find the general and singular solutions (if any) of…

Find the general and singular solutions (if any) of $y(y-2xy')^2=2y'$.

Step 1: Expand and confirm the quadratic in $p$

Expanding: $y^3-4xy^2p+4x^2yp^2-2p=0$, i.e. \[4x^2y\,p^2-(4xy^2+2)p+y^3=0,\] matching the source (sympy-confirmed). Its discriminant is $4(4xy^2+1)$, giving \[p=\frac{(2xy^2+1)\pm\sqrt{4xy^2+1}}{4x^2y}.\]

Step 2: $p$-discriminant $\Rightarrow$ singular solution

$F_p=2\big(4x^2yp-2xy^2-1\big)=0\Rightarrow p=\dfrac{2xy^2+1}{4x^2y}$; substituting into $F=0$ gives $F=-\dfrac{4xy^2+1}{4x^2y}=0\Rightarrow xy^2=-\tfrac14$. On this curve, $x=-1/(4y^2)$, so $dx/dy=1/(2y^3)\Rightarrow y'=2y^3$; substituting into the original equation, sympy confirms $y(y-2xy')^2-2y'\equiv 0$. This matches the source and is a genuine singular solution.

Step 3: Isobaric substitution reduces the equation

Assigning weight $1$ to $x$ and $w$ to $y$, every term of $y(y-2xy')^2-2y'$ has total weight $3w$ on the left provided $y'$ carries weight $w-1$; matching against the right side ($2y'$, weight $w-1$) forces $3w=w-1\Rightarrow w=-\tfrac12$. Substituting $y=v(x)\,x^{-1/2}$ and clearing $x^{3/2}$, sympy reduces the equation to \[4x^2v\,(v')^2-2x(4v^2+1)v'+(4v^2+1)v=0,\] a quadratic in $v'=dv/dx$ with $x$-dependence only through explicit $x$ factors (no leftover fractional powers) — confirming the isobaric reduction.

Step 4: Solve the quadratic for $v'$ and separate

With $s=\sqrt{4v^2+1}$, the quadratic formula gives $v'=\dfrac{s(s\pm1)}{4xv}$. Using $s^2=4v^2+1\Rightarrow v\,dv=\tfrac14 s\,ds$, this separates cleanly: \[\frac{dx}{x}=\frac{4v\,dv}{s(s\pm1)}=\frac{ds}{s\pm1}.\]

Step 5: Integrate and eliminate the parameter

Integrating: $\ln x=\ln(s\pm1)+\text{const}\Rightarrow x=A(s\pm1)$, i.e. $s=\dfrac{x}{A}\mp1$. Squaring $s^2=4v^2+1$ and using $v^2=xy^2$ (since $v=y\sqrt x$), dividing by $x$, and simplifying (sympy) collapses the $\pm$ ambiguity into a single one-parameter family after renaming the constant $C=1/(2A)$: \[y^2=C^2x+C.\]

Step 6: Verify the general solution directly

With $y=\sqrt{C^2x+C}$, sympy computes $y'$ and substitutes into $y(y-2xy')^2-2y'$: the residual simplifies exactly to $0$ for arbitrary $C$ — a genuine, elementary, closed-form general solution (the source never got past the unintegrated quadratic-in-$p$ expression).

Step 7: Cross-check the singular solution as the envelope

The envelope of $y^2-C^2x-C=0$ satisfies $\partial_C(\cdots)=0\Rightarrow -2Cx-1=0\Rightarrow C=-\dfrac1{2x}$; substituting back gives $y^2=-\dfrac1{4x}$, i.e. $xy^2=-\tfrac14$ — exactly the singular solution found independently via the $p$-discriminant in Step 2. This full consistency confirms both results.