Solve (y')^2-(x+y)y'+xy=0 for y' and integrate
Solve $(y')^2-(x+y)y'+xy=0$ for $y'$ and integrate.
Explanation of Method
Factor the quadratic in $p=y'$ directly into two linear factors and solve each branch; confirm the factorization is not itself a source of a singular solution.
Step 1: Factor the quadratic in $p$
$p^2-(x+y)p+xy=(p-x)(p-y)=0$ (verified by expansion with sympy), so $p=x$ or $p=y$.
Step 2: Solve branch $p=x$
$y'=x\ \Rightarrow\ y=\dfrac{x^2}{2}+C_1$.
Step 3: Solve branch $p=y$
$y'=y\ \Rightarrow\ y=C_2e^{x}$.
Step 4: Check for a singular solution along $y=x$
The two roots $p=x$ and $p=y$ coincide only where $x=y$. Substituting $y=x,\ y'=1$ into the original equation gives \[p^2-(x+y)p+xy=1-2x+x^2=(x-1)^2,\] not identically $0$, so $y=x$ does not satisfy the ODE and there is no singular solution.
Result
Final answer:
\[\boxed{\,y=\frac{x^2}{2}+C_1\ \ \text{or}\ \ y=C_2e^{x}\,}\]
No singular solution exists.
Editor's note: this problem is a verbatim duplicate of Problem 160 in the archive — same equation, same solution, reproduced here with the same verified treatment (factorization and both residual checks confirmed independently with sympy).