Solve y=y'^5+y'^3+5 using the parameter method
Solve $y=y'^5+y'^3+5$ using the parameter method.
Explanation of Method
With $y$ given directly in terms of $p=y'$, differentiate to get $dy/dp$, use $dy=p\,dx$ to solve for $dx/dp$, then integrate.
Step 1: Set $p=y'$
\[y=p^5+p^3+5.\]
Step 2: Differentiate (sympy)
\[\frac{dy}{dp}=5p^4+3p^2,\]
confirmed with sympy's diff.
Step 3: Use $dy=p\,dx$ to solve for $dx/dp$
\[\frac{dx}{dp}=\frac{1}{p}\frac{dy}{dp}=\frac{5p^4+3p^2}{p}=5p^3+3p,\] sympy simplifies the quotient directly to this form, confirming the source.
Step 4: Integrate
\[x=\int(5p^3+3p)\,dp+C=\frac{5p^4}{4}+\frac{3p^2}{2}+C,\]
verified with sympy's integrate.
Step 5: End-to-end verification
Computing $\dfrac{dy}{dx}=\dfrac{dy/dp}{dx/dp}=\dfrac{5p^4+3p^2}{5p^3+3p}$ and simplifying with sympy gives exactly $p$, identically — confirming $\{x(p),y(p)\}$ is a genuine solution curve of $y=y'^5+y'^3+5$.
Result
Final answer:
\[\boxed{\,x=\frac{5p^4}{4}+\frac{3p^2}{2}+C,\qquad y=p^5+p^3+5\,}\]
Every derivative, the algebraic simplification of $dx/dp$, the integration, and the $dy/dx=p$ end-to-end check were verified with sympy. The source's solution is correct.