Solve y=y'^2-y'x+(x^2)/(2) using the parameter method, and…

Solve $y=y'^2-y'x+\dfrac{x^2}{2}$ using the parameter method, and determine whether the candidate curve $y=x^2/4$ is a genuine singular solution.

Step 1: Implicit differentiation for $p(x)$

With $p=y'$, $y=p^2-px+x^2/2$. Differentiating symbolically in \texttt{sympy} (treating $p=p(x)$) and setting $dy/dx=p$ gives, after simplification, \[(2p-x)\,p'(x)=2p-x.\] So, provided $2p-x\ne0$, $p'(x)=1$.

Step 2: Integrate

$p'(x)=1\Rightarrow p=x+\text{const}$, i.e. $x=p+C$ — matching the source.

Step 3: Back-substitute

Substituting $x=p+C$ into $y=p^2-px+x^2/2$ and simplifying gives $y=\dfrac{p^2}{2}+\dfrac{C^2}{2}$ (\texttt{sympy}-confirmed), matching the source. Eliminating $p=x-C$: \[y=\frac{x^2}{2}-Cx+C^2.\]

Step 4: Direct-substitution verification of the general solution

With $y=\dfrac{x^2}{2}-Cx+C^2$, $y'=x-C=p$. \texttt{sympy} confirms \[y-(p^2-px+x^2/2)\equiv 0\] identically in $x,C$ — the general family is correct.

Step 5: Proper $C$-discriminant test for the envelope

Let $G(x,y,C)=y-\dfrac{x^2}{2}+Cx-C^2$. Then \[\frac{\partial G}{\partial C}=x-2C=0 \implies C=\frac{x}{2}.\] Substituting back: $G\left(x,y,\tfrac x2\right)=y-\dfrac{x^2}{4}$, so the $C$-discriminant elimination gives exactly $y=\dfrac{x^2}{4}$.

Step 6: Confirm the envelope solves the ODE

With $y=x^2/4$, $y'=x/2=p$. \texttt{sympy} confirms $y-(p^2-px+x^2/2)\equiv 0$. So $y=x^2/4$ is not just a curve that happens to satisfy the equation — it is genuinely and rigorously the envelope of the family $y=\tfrac{x^2}{2}-Cx+C^2$, confirmed by the standard $C$-discriminant test (not merely speculated, as the source did).