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

Solve $y=2xy'+\frac{x^2}{2}+(y')^2$ using the parameter method, and find both the general and singular solutions.

Step 1: Differentiate and split into branches

Let $p=y'$. Differentiating $y=2xp+\tfrac{x^2}{2}+p^2$ and imposing $dy/dx=p$ with \texttt{sympy}: $p=2p+x+p'(2x+2p)$, so $2p'(x+p)=-(x+p)$. For $x+p\ne0$ this gives $p'=-\tfrac12$, hence $p=-\tfrac{x}{2}+C$ — matching the source's Step 4.

Step 2: Independently re-expand the general solution (sympy)

Rather than checking the source's hand algebra step-by-step, $p=C-\tfrac{x}{2}$ is substituted into $y=2xp+\tfrac{x^2}{2}+p^2$ and expanded from scratch with \texttt{sympy.expand}, giving

identical to the source's boxed $y=Cx-\tfrac{x^2}{4}+C^2$. \texttt{sympy} further confirms $y'=C-\tfrac{x}{2}=p$ exactly, and direct substitution of $y,y'$ into $y-\big(2xy'+\tfrac{x^2}{2}+(y')^2\big)$ returns $0$ identically — the general solution and the source's Step 7 checkmark are both correct.

Step 3: Singular solution from $x+p=0$

The excluded branch $x+p=0$, i.e. $p=-x$, substituted into $y=2xp+\tfrac{x^2}{2}+p^2$ and simplified independently with \texttt{sympy} gives

matching the source's Step 8. Direct substitution ($y'=-x$) into $y-\big(2xy'+\tfrac{x^2}{2}+(y')^2\big)$ returns $0$ identically — the singular solution and the source's Step 8 checkmark are both correct.