Solve y=x(1+y')+(y')^2 using the parameter method, and…
Solve $y=x(1+y')+(y')^2$ using the parameter method, and determine whether a singular solution exists.
Explanation of Method
Parameter substitution $p=y'$, reduction to a linear ODE for $x(p)$ solved with integrating factor $e^p$; the source's confused singular-solution derivation is discarded and redone rigorously via the $p$-discriminant, then checked directly against the original equation with \texttt{sympy}.
Step 1: Linear ODE for $x(p)$
Let $p=y'$. Differentiating $y=x+xp+p^2$ and imposing $dy/dx=p$ with \texttt{sympy}: $p=1+p+p'(x+2p)$, so $p'(x+2p)=-1$. Inverting gives the linear ODE, matching the source exactly:
\[ \frac{dx}{dp}+x=-2p \]Step 2: Solve and verify the general solution (sympy)
\texttt{sympy.dsolve} (integrating factor $e^p$) gives $x=2-2p+Ce^{-p}$. Substituting into $y=x(1+p)+p^2$ and expanding independently with \texttt{sympy} (not merely checking the source's hand algebra) gives
\[ y=2-p^2+Ce^{-p}(1+p) \]exactly matching the source. The gold-standard check $\dfrac{dy/dp}{dx/dp}-p$ simplifies to $0$, confirming this general solution is correct.
Step 3: Why the source's singular-solution derivation is unusable
The source's intermediate step ("$dy/dp=dx/dp+x+p\,dx/dp+2p=0$") is algebraically incoherent and is discarded. The correct $p$-discriminant procedure is: from $F(x,y,p)=y-x-xp-p^2=0$, $\partial F/\partial p=-x-2p=0$ gives $x=-2p$; substituting into $y=x+xp+p^2$ gives $y=-2p-p^2$; eliminating $p=-x/2$ gives $y=x-\tfrac14x^2$ — which happens to match the source's final boxed answer despite the broken intermediate algebra.
Step 4: Direct substitution shows this candidate is wrong (sympy)
Substituting $y=x-\tfrac14x^2$ (so $y'=1-\tfrac{x}{2}$) into $y-\big(x(1+y')+(y')^2\big)$ and simplifying with \texttt{sympy} gives a residual of $\mathbf{-1}$, not $0$ — the candidate is off by an additive constant and is not a solution of the original equation. This is confirmed independently: the Lagrange fixed-point test $p=\varphi(p)$ with $\varphi(p)=1+p$ gives $p=1+p\Rightarrow0=1$, which has \emph{no} solution, so no envelope-type line solution exists either. A further check using the Jacobian $\partial(x,y)/\partial(p,C)$ of the verified general-solution family shows it vanishes only along this same spurious curve $x=-2p,\ y=-p^2-2p$ — and \texttt{sympy} confirms the family's own slope there is $p$ while the candidate curve's own slope is $p+1$, i.e. the two are never tangent. So this is a spurious locus, not a genuine envelope.
Result
Final answer:
General solution verified correct; source's singular solution is wrong and does not exist.
\[ x=2-2p+Ce^{-p},\qquad y=2-p^2+Ce^{-p}(1+p) \]is the complete, sympy-verified general solution (gold-standard ratio test vanishes identically). The source's boxed singular solution $y=x-\tfrac14x^2$ — despite arriving at the same final formula as a correct $p$-discriminant elimination — fails direct substitution into the original equation (\texttt{sympy} residual $=-1\ne0$), so it is not a valid solution. Both the Lagrange fixed-point test ($p=1+p$ has no solution) and a Jacobian/tangency check on the general-solution family confirm this equation has no singular solution; the "singular solution" in the source should be deleted entirely.