Solve (y')^2-(x+y)y'+xy=0 for y' and integrate to find the…
Solve $(y')^2-(x+y)y'+xy=0$ for $y'$ and integrate to find the general and singular solutions (if any).
Explanation of Method
Factor the quadratic in $p=y'$ (its roots are simply $p=x$ and $p=y$), integrate each branch, then apply the $p$-discriminant test to check for a singular solution.
Step 1: Factor the quadratic in $p$
With $p=y'$: $p^2-(x+y)p+xy=(p-x)(p-y)$ (sympy-confirmed by expansion), so $p=x$ or $p=y$. The discriminant is $(x+y)^2-4xy=(x-y)^2$, a perfect square, matching the source.
Step 2: Integrate each branch
Branch $p=x$: $y'=x\Rightarrow y=\dfrac{x^2}2+C_1$.
Branch $p=y$: $y'=y\Rightarrow y=C_2e^{x}$.
Sympy confirms both satisfy the original equation identically (residual $0$ for each, substituting $y$ and $y'$ back into $(y')^2-(x+y)y'+xy$).
Step 3: $p$-discriminant test
$F=p^2-(x+y)p+xy$, $F_p=2p-(x+y)=0\Rightarrow p=\dfrac{x+y}2$; substituting back, sympy gives $F=-\dfrac{(x-y)^2}4=0\Rightarrow x=y$, i.e. candidate $y=x$.
Step 4: Test the candidate $y=x$
With $y=x$, $y'=1$: substituting into $(y')^2-(x+y)y'+xy$ gives $1-2x+x^2=(x-1)^2$ (sympy-confirmed), which is not identically $0$ (only vanishes at the single point $x=1$). Hence $y=x$ is not a solution curve, and there is no singular solution — matching the source's conclusion.
Result
Final answer:
General solution (two independent families): \[\boxed{\,y=\frac{x^2}{2}+C_1\,}\qquad\text{or}\qquad\boxed{\,y=C_2e^{x}\,}\] No singular solution exists (the $p$-discriminant candidate $y=x$ fails to satisfy the original equation identically). All algebra, the discriminant, and both verification substitutions were checked symbolically with sympy; the source's analysis here is fully correct.