Solve…

Solve $(y')^3-(x^2+xy+y^2)(y')^2+(x^3y+x^2y^2+xy^3)y'-x^3y^3=0$ for $y'$ and integrate each branch.

Step 1: Verify the cubic factorization

Expanding $(p-xy)(p-x^2)(p-y^2)$ with sympy gives \[p^3-(x^2+xy+y^2)p^2+(x^3y+x^2y^2+xy^3)p-x^3y^3,\] term-for-term identical to the given cubic (the symbolic difference simplifies to $0$) — the factorization is confirmed exact.

Step 2: Solve branch $p=xy$

$y'=xy\ \Rightarrow\ \dfrac{dy}{y}=x\,dx\ \Rightarrow\ y=C_1e^{x^2/2}$ (confirmed by direct substitution).

Step 3: Solve branch $p=x^2$

$y'=x^2\ \Rightarrow\ y=\dfrac{x^3}{3}+C_2$ (confirmed by direct substitution).

Step 4: Solve branch $p=y^2$

$y'=y^2\ \Rightarrow\ -\dfrac1y=x+C\ \Rightarrow\ y=-\dfrac{1}{x+C_3}$ (confirmed by direct substitution).

Step 5: $p$-discriminant / singular-solution check

The three roots $p=xy,\,x^2,\,y^2$ coincide along $y=x$ (triple root $p=x^2$) and along $y=-x$. Testing $y=x$ with its actual derivative $y'=1$ in the original cubic gives residual $-(x^2-1)^3\ne0$ identically; testing $y=-x$ with $y'=-1$ gives residual $(x^2-1)(x^2+1)^2\ne0$ identically (both confirmed with sympy). Neither curve actually satisfies the ODE, so there is no singular solution.