Find the general and singular solutions of x(y')^3+x=1

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

Step 1: Solve for $x$ in terms of $p$

$x(p^3+1)=1\Rightarrow x=\dfrac{1}{p^3+1}$.

Step 2: Build $dy/dp$

$dy=p\,dx$, and $\dfrac{dx}{dp}=\dfrac{-3p^2}{(p^3+1)^2}$, so \[\frac{dy}{dp}=p\cdot\frac{dx}{dp}=\frac{-3p^3}{(p^3+1)^2}.\]

Step 3: Correct integration (this is where the source stalled)

Writing $p^3=(p^3+1)-1$: $\dfrac{-3p^3}{(p^3+1)^2}=\dfrac{-3}{p^3+1}+\dfrac{3}{(p^3+1)^2}$. Carrying out the partial-fraction decomposition of $1/(p^3+1)=1/[(p+1)(p^2-p+1)]$ and integrating (independently verified with symbolic integration — the derivative of the result below matches $dy/dp$ identically, residual $\equiv0$) gives the closed form \[y(p)=\frac{p}{p^3+1}-\frac13\ln(p+1)+\frac16\ln(p^2-p+1)-\frac{1}{\sqrt3}\arctan\!\left(\frac{2p-1}{\sqrt3}\right)+C.\] (The source's unverified guess $y=p/(p^3+1)+C$ is incomplete — it is missing exactly these log/arctan terms, which is why the source's own derivative check failed to reproduce $p$; that failure was real and never resolved in the source.)

Step 4: Verify the corrected parametric solution

Direct symbolic check confirms $\dfrac{dy/dp}{dx/dp}=p$ identically, i.e. the parametric pair \[x=\frac{1}{p^3+1},\qquad y=\frac{p}{p^3+1}-\frac13\ln(p+1)+\frac16\ln(p^2-p+1)-\frac{1}{\sqrt3}\arctan\!\left(\frac{2p-1}{\sqrt3}\right)+C\] satisfies $dy/dx=p$, and hence $x(y')^3+x-1\equiv0$ along the curve. This is the correctly derived general solution.

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

$F=xp^3+x-1$, $F_p=3xp^2=0\Rightarrow x=0$ or $p=0$. If $x=0$: $F=-1\neq0$ (no solution). If $p=0$: $F=x-1=0\Rightarrow x=1$. So the only candidate is the vertical line $x=1$, occurring at parameter $p=0$.

But at $p=0$, $x=1$ for every choice of $C$ — it is simply the point where each member of the general family has a horizontal tangent ($p=0$), with $y$-value $C+\pi/(6\sqrt3)$ depending on $C$. As $C$ ranges over $\mathbb R$, these points sweep out the entire line $x=1$, which is therefore already fully contained in (traced out by) the general solution. It is a tac/node-type locus of horizontal-tangent points, not an independent envelope curve.