Solve (y')^2+yy'-x(x+y)=0 for y' and integrate each…

Solve $(y')^2+yy'-x(x+y)=0$ for $y'$ and integrate each resulting branch; check for a singular solution.

Step 1: Quadratic in $p$ and its discriminant

$p^2+yp-x(x+y)=0$ is quadratic in $p$. Its discriminant is \[\Delta=y^2+4x(x+y)=y^2+4xy+4x^2=(y+2x)^2\ \ (\ge 0),\] confirmed symbolically, so \(p=\dfrac{-y\pm(y+2x)}{2}\), giving the two branches $p=x$ and $p=-x-y$.

Step 2: Solve branch $p=x$

$y'=x\ \Rightarrow\ y=\dfrac{x^2}{2}+C_1$.

Step 3: Solve branch $p=-x-y$

This is the linear ODE $y'+y=-x$. With integrating factor $e^x$: $(e^xy)'=-xe^x$, so \[e^xy=\int -xe^x\,dx=(1-x)e^x+C,\] giving $y=1-x+Ce^{-x}$ (confirmed with sympy's dsolve).

Step 4: $p$-discriminant test

With $F=p^2+yp-x^2-xy$: $F_p=2p+y=0\Rightarrow p=-y/2$; substituting into $F=0$ gives $(y+2x)^2=0\Rightarrow y=-2x$.

Step 5: Check $y=-2x$ against the original equation

With $y=-2x,\ y'=-2$: \[(y')^2+yy'-x(x+y)=4+4x+x^2=(x+2)^2,\] which is not identically $0$. So $y=-2x$ does not satisfy the ODE in general and is not a singular solution.