Solve y=2xy'-(y')^3 using the parameter method, and…
Solve $y=2xy'-(y')^3$ using the parameter method, and determine the true singular solution.
Explanation of Method
Parameter substitution $p=y'$, reduction to a linear ODE for $x(p)$; the source's $p$-discriminant singular-solution candidate is checked directly against the original ODE with \texttt{sympy} rather than trusted as an envelope (this is a Lagrange-type equation, not a true Clairaut equation, since the coefficient of $x$ is $2p$, not $p$).
Step 1: Linear ODE for $x(p)$
Let $p=y'$. Differentiating $y=2xp-p^3$ and imposing $dy/dx=p$ with \texttt{sympy} gives $p=2p+p'(2x-3p^2)$, i.e. $p'(2x-3p^2)=-p$. Inverting ($dx/dp=1/p'$) reproduces the source's linear equation exactly:
\[ \frac{dx}{dp}+\frac{2x}{p}=3p \]Step 2: Solve and verify the general solution (sympy)
\texttt{sympy.dsolve} with integrating factor $p^2$ gives $x=\dfrac{3p^2}{4}+\dfrac{C}{p^2}$, and back-substituting into $y=2xp-p^3$ gives $y=\dfrac{p^3}{2}+\dfrac{2C}{p}$ — matching the source exactly. The gold-standard check, $\dfrac{dy/dp}{dx/dp}-p$, simplifies to $0$ under \texttt{sympy}, and direct substitution of $x(p),y(p)$ into $y-(2xp-p^3)$ also returns $0$ identically. The general solution is fully correct.
Step 3: The source's $p$-discriminant candidate, and why it fails
From $F(x,y,p)=y-2xp+p^3=0$, $\partial F/\partial p=-2x+3p^2=0$ gives $x=\tfrac{3}{2}p^2$, and back-substituting gives $y=2p^3$ — reproducing the source's computation of $27y^2=32x^3$ exactly. But the source never checks this against the original equation. Using \texttt{sympy} to compute the \emph{actual} slope along this candidate curve, $\dfrac{dy/dp}{dx/dp}=2p$, not $p$ — already a red flag, since a genuine envelope must be tangent to the family (slope $p$) at every point. Substituting the actual slope $2p$ back into $y-(2x\cdot(2p)-(2p)^3)$ gives a residual of $4p^3\not\equiv0$. An independent explicit check (writing $y=\sqrt{32x^3/27}$ and its true derivative) gives the same nonzero residual, $\dfrac{8\sqrt6}{9}x^{3/2}$. So $27y^2=32x^3$ is not a solution of the original ODE — it is a spurious ("tac-locus"-type) curve produced by the $p$-discriminant of $F$, which is not a reliable envelope test for this non-Clairaut, Lagrange-type equation (the same phenomenon documented for Problem 180).
Step 4: The true singular solution (Lagrange fixed-point test)
Writing the equation in Lagrange form $y=x\varphi(p)+\psi(p)$ with $\varphi(p)=2p$, $\psi(p)=-p^3$, the genuine singular (envelope) solutions of a Lagrange equation come from the fixed points $p=\varphi(p)$, i.e. $p=2p\Rightarrow p=0$ — not from $\partial F/\partial p=0$. Substituting $p=0$ gives the candidate line $y=x\varphi(0)+\psi(0)=0$. Direct \texttt{sympy} substitution of $y\equiv0$ (so $y'\equiv0$) into $y-(2xy'-(y')^3)$ returns $0$ identically. This is the genuine singular solution.
Result
Final answer:
General solution verified correct; source's singular solution is wrong and is replaced.
\[ x=\frac{3p^2}{4}+\frac{C}{p^2},\qquad y=\frac{p^3}{2}+\frac{2C}{p} \]is the complete, sympy-verified general solution (gold-standard ratio test and direct substitution both vanish identically). The source's boxed singular solution $27y^2=32x^3$, obtained from the $p$-discriminant of $F(x,y,p)$, was never checked against the original equation by the source — and indeed \texttt{sympy} shows it leaves a nonzero residual ($4p^3$, equivalently $\tfrac{8\sqrt6}{9}x^{3/2}$) when substituted back, so it is not a valid solution. The genuine singular solution, found correctly via the Lagrange fixed-point condition $p=\varphi(p)$ (here $p=2p\Rightarrow p=0$) and confirmed by direct substitution, is simply
\[ y=0 \]