Solve xy'^3-1-y'=0 using the parameter method

Solve $xy'^3-1-y'=0$ using the parameter method.

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

With $p=y'$: $xp^3=1+p\ \Rightarrow\ x=\dfrac{1+p}{p^3}$.

Step 2: Differentiate (sympy)

\[\frac{dx}{dp}=-\frac{2p+3}{p^4}\] computed independently with sympy's diff, confirming the source's expression.

Step 3: Build $dy=p\,dx$

\[\frac{dy}{dp}=p\cdot\frac{dx}{dp}=-\frac{2p+3}{p^3}=-\left(\frac{2}{p^2}+\frac{3}{p^3}\right),\] which sympy confirms matches the source's claimed form exactly.

Step 4: Integrate

\[y=\int\left(-\frac{2}{p^2}-\frac{3}{p^3}\right)dp+C=\frac{2}{p}+\frac{3}{2p^2}+C,\] verified with sympy's integrate — matches the source.

Step 5: End-to-end verification

Two independent checks confirm this is a genuine solution curve: (i) $x(p)\,p^3-1-p=0$ identically (holds by construction); (ii) computing $\dfrac{dy}{dx}=\dfrac{dy/dp}{dx/dp}$ from the parametric pair gives exactly $p$, identically in $p$ — the gold-standard confirmation that $\{x(p),y(p)\}$ solves the original equation.