Solve y=2xy'-(y')^2 using the parameter method, and…
Solve $y=2xy'-(y')^2$ using the parameter method, and determine whether a singular solution exists.
Explanation of Method
Parameter substitution $p=y'$; this is \emph{not} a true Clairaut equation (the coefficient $2\ne1$ on $xy'$ breaks the Clairaut structure), so it is treated as a Lagrange equation, and any $p$-discriminant candidate is checked directly against the original ODE with \texttt{sympy} rather than trusted as an envelope.
Step 1: Why this is not a Clairaut equation
A true Clairaut equation $y=xp+f(p)$ differentiates to $p'\cdot(x+f'(p))=0$, giving a free branch $p'=0$ (arbitrary constant $p=C$) plus an envelope. Here $y=2xp-p^2$ differentiates (via \texttt{sympy}, imposing $dy/dx=p$) to
\[ p=2p+p'(2x-2p) \quad\Longrightarrow\quad p'=\frac{-p}{2x-2p} \]The right-hand side is $-p$, not $0$: setting $p'=0$ forces $p=0$ specifically, not an arbitrary constant. So the coefficient $2$ genuinely breaks the Clairaut structure, and this must be solved as a Lagrange equation instead — confirming the source's own observation.
Step 2: Linear ODE for $x(p)$
Inverting $dp/dx=\dfrac{-p}{2x-2p}$ with \texttt{sympy} gives $dx/dp=2-2x/p$, i.e. the linear equation
\[ \frac{dx}{dp}+\frac{2x}{p}=2 \]reproducing the source's Step 5 exactly.
Step 3: Solve and verify the general solution (sympy)
\texttt{sympy.dsolve} on this linear ODE (equivalently, using the integrating factor $p^2$) returns
\[ x=\frac{2p}{3}+\frac{C}{p^2} \]Substituting back, $y=2xp-p^2=\dfrac{p^2}{3}+\dfrac{2C}{p}$. The gold-standard check — \texttt{sympy} computing $\dfrac{dy/dp}{dx/dp}$ — simplifies identically to $p$, and direct substitution of $x(p),y(p)$ into $y-(2xp-p^2)$ returns $0$ identically. The general solution (source's Steps 5–7) is fully correct.
Step 4: $p$-discriminant candidate and its failure (sympy)
From $F(x,y,p)=2xp-p^2-y=0$, $\partial F/\partial p=2x-2p=0$ gives $p=x$, and substituting into the original equation gives the candidate $y=2x\cdot x-x^2=x^2$ — reproducing the source's $p$-discriminant computation exactly. But substituting $y=x^2$ (so $y'=2x$) into the original equation and simplifying with \texttt{sympy}:
\[ y-(2xy'-(y')^2) = x^2-(4x^2-4x^2)=x^2 \]This residual is $x^2$, not identically zero (it vanishes only at the single point $x=0$). The source's own Step 9 shows exactly this contradiction — and then still boxes $y=x^2$ as "the singular solution" despite having just disproved it. That is a self-contradictory error in the source.
Result
Final answer:
General solution verified correct; source's "singular solution" is corrected (removed).
\[ x=\frac{2p}{3}+\frac{C}{p^2},\qquad y=\frac{p^2}{3}+\frac{2C}{p} \]is the complete, sympy-verified general solution (gold-standard ratio test and direct substitution both vanish identically). The candidate $y=x^2$ obtained from the $p$-discriminant is not a genuine solution: substituting it back into $y=2xy'-(y')^2$ leaves a residual of $x^2\not\equiv0$, exactly as the source's own unfinished check already showed before it was (incorrectly) boxed as an answer. Since this equation is not truly Clairaut, there is no reason to expect a free-constant envelope branch in the first place, and none survives verification: this equation has no singular solution — the boxed $y=x^2$ in the source should be deleted, leaving only the parametric general solution above.