Solve x=y'^3-y'-1 using the parameter method
Solve $x=y'^3-y'-1$ using the parameter method.
Explanation of Method
With $x$ given directly in terms of $p=y'$, differentiate, use $dy=p\,dx$, integrate, and locate the critical points where $dx/dp=0$.
Step 1: Set $p=y'$
\[x=p^3-p-1.\]
Step 2: Differentiate (sympy)
\[\frac{dx}{dp}=3p^2-1,\]
confirmed with sympy's diff.
Step 3: Build $dy=p\,dx$
\[dy=p(3p^2-1)\,dp=(3p^3-p)\,dp,\]
the multiplication is confirmed symbolically with sympy's expand.
Step 4: Integrate
\[y=\int(3p^3-p)\,dp+C=\frac{3p^4}{4}-\frac{p^2}{2}+C,\]
verified with sympy's integrate — matches the source.
Step 5: Critical points ($dx/dp=0$)
Solving $3p^2-1=0$ with sympy gives $p=\pm\dfrac{1}{\sqrt3}=\pm\dfrac{\sqrt3}{3}$. Substituting back into $x=p^3-p-1$ (sympy, exact and numeric): \[p=\tfrac{\sqrt3}{3}:\ x=-1-\tfrac{2\sqrt3}{9}\approx-1.385,\qquad p=-\tfrac{\sqrt3}{3}:\ x=-1+\tfrac{2\sqrt3}{9}\approx-0.615.\] These match the source's claimed critical values exactly.
Step 6: End-to-end verification
Computing $\dfrac{dy}{dx}=\dfrac{dy/dp}{dx/dp}=\dfrac{3p^3-p}{3p^2-1}$ and simplifying with sympy gives exactly $p$, identically — confirming the parametric pair solves $x=y'^3-y'-1$.
Result
Final answer:
\[\boxed{\,x=p^3-p-1,\qquad y=\frac{3p^4}{4}-\frac{p^2}{2}+C\,}\]
Critical points at $p=\pm\sqrt3/3$ give $x\approx-1.385$ and $x\approx-0.615$ (both verified numerically). All derivatives, the integration, and the $dy/dx=p$ end-to-end check were verified with sympy. The source's solution, including the critical-point computation, is correct.