Find the general and singular solutions of…

Find the general and singular solutions of $(y')^3+y^2=yy'(y'+1)$.

Step 1: Expand and factor

$p^3+y^2=yp(p+1)=yp^2+yp \Rightarrow p^3-yp^2-yp+y^2=0$. Symbolically expanding $(p-y)(p^2-y)=p^3-yp^2-py+y^2$, which matches the left side exactly (verified). So the equation factors as $(p-y)(p^2-y)=0$.

Step 2: Solve branch 1, $y'=y$

$y=C_1e^{x}$. Substitution: $y'=C_1e^x=y$, so $(y')^3+y^2-yy'(y'+1)=y^3+y^2-y\cdot y\cdot(y+1)=y^3+y^2-y^3-y^2=0$ identically (confirmed symbolically for all $C_1$).

Step 3: Solve branch 2, $(y')^2=y$

Separating, $y=\dfrac{(x+C_2)^2}{4}$. Substitution confirms the residual $(y')^3+y^2-yy'(y'+1)\equiv0$ symbolically for all $C_2$.

Step 4: $p$-discriminant, checked carefully

$F=p^3-yp^2-yp+y^2$, $F_p=3p^2-2py-y$. Solving $F=0,\,F_p=0$ simultaneously gives two algebraic roots: $(p,y)=(0,0)$ and $(p,y)=(1,1)$.

$(p,y)=(0,0)$: corresponds to the constant curve $y\equiv0$, which indeed has $y'=0=p$ — self-consistent, and it is already the member $C_1=0$ of branch 1.

$(p,y)=(1,1)$: would require the constant curve $y\equiv1$, whose actual derivative is $y'=0$, not $p=1$ as the root demands — substituting $y=1$ into the original cubic $p^3-p^2-p+1=(p-1)^2(p+1)=0$ gives roots $p=1,1,-1$, none of which is $p=0$. So $y\equiv1$ does not satisfy the ODE (its true derivative $0$ is not a root). This candidate is a spurious algebraic artifact of solving the discriminant system without the constraint $p=y'$ for a constant curve, and must be discarded.