Solve x^3+y'^3-3xy'=0 using the parameter method

Solve $x^3+y'^3-3xy'=0$ using the parameter method.

Step 1: Set $p=y'$ and substitute $x=tp$

The equation is $x^3+p^3-3xp=0$ (a folium-of-Descartes-type curve in $x,p$). Setting $x=tp$ (i.e. $t=x/p$) and expanding with sympy: \[p^3t^3+p^3-3p^2t=0\ \Longrightarrow\ p^2\big(p(t^3+1)-3t\big)=0.\] The factor $p^2=0$ only gives the trivial point $p=0,\,x=0$; the genuine branch is \[p(t^3+1)=3t\ \Longrightarrow\ p(t)=\frac{3t}{t^3+1}.\]

Step 2: Recover $x(t)$

\[x(t)=t\,p(t)=\frac{3t^2}{t^3+1}.\] Sanity check with sympy: substituting $x(t),p(t)$ back into $x^3+p^3-3xp$ simplifies to $0$ identically — the substitution is valid.

Step 3: Differentiate $x(t)$ (sympy)

\[\frac{dx}{dt}=\frac{3t(2-t^3)}{(t^3+1)^2}\] (sympy simplifies the raw quotient-rule denominator $t^6+2t^3+1$ to $(t^3+1)^2$).

Step 4: Build $dy/dt=p\cdot dx/dt$

\[\frac{dy}{dt}=\frac{3t}{t^3+1}\cdot\frac{3t(2-t^3)}{(t^3+1)^2}=\frac{9t^2(2-t^3)}{(t^3+1)^3}.\]

Step 5: Integrate (sympy, closed form)

Sympy's integrate finds a clean closed form directly: \[y(t)=\frac{3(2t^3-1)}{2(t^3+1)^2}+C.\] (Differentiating this back with sympy reproduces $dy/dt$ exactly, confirming the antiderivative.) This replaces the source's incomplete, unresolved partial-fraction attempt with a single verified closed-form expression.

Step 6: End-to-end verification

Computing $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ from the parametric pair and simplifying with sympy gives exactly $p(t)=\dfrac{3t}{t^3+1}$, identically in $t$ — the definitive confirmation that $\{x(t),y(t)\}$ is a genuine solution curve of $x^3+y'^3-3xy'=0$.