Solve x-y=(4)/(9)y'^2-(8)/(27)y'^3 using the parameter…

Solve $x-y=\dfrac{4}{9}y'^2-\dfrac{8}{27}y'^3$ using the parameter method $p=y'$.

Step 1: Rewrite and differentiate w.r.t.\ $y$

With $p=y'$, $x=y+\dfrac{4}{9}p^2-\dfrac{8}{27}p^3$. Using $\dfrac{dx}{dy}=\dfrac1p$, \texttt{sympy} differentiation of the right side (with $p=p(y)$) and equating to $1/p$ gives \[\frac1p=1+\left(\frac89p-\frac89p^2\right)\frac{dp}{dy}=1+\frac89p(1-p)\frac{dp}{dy},\] matching the source's setup exactly.

Step 2: Isolate $dy/dp$ (for $p\ne1$)

$\dfrac1p-1=\dfrac{1-p}p=\dfrac89p(1-p)\dfrac{dp}{dy}$. For $p\ne1$ this cancels to $\dfrac1p=\dfrac89p\,\dfrac{dp}{dy}$, i.e. \[\frac{dy}{dp}=\frac89p^2.\]

Step 3: Integrate and back-substitute

$y=\dfrac{8}{27}p^3+C$. Substituting into $x=y+\dfrac49p^2-\dfrac{8}{27}p^3$, \texttt{sympy} simplifies the $p^3$ terms away exactly, leaving \[x=C+\frac49p^2,\qquad y=C+\frac{8}{27}p^3.\]

Step 4: Gold-standard verification

\texttt{sympy} computes $\dfrac{dy}{dp}=\dfrac89p^2$, $\dfrac{dx}{dp}=\dfrac89p$, and confirms \[\frac{dy/dp}{dx/dp}-p\] simplifies to exactly $0$ — the parametric general solution is genuine.

Step 5: Singular solution at $p=1$

The division by $1-p$ in Step 2 fails at $p=1$. Substituting $p=1$ directly into the original relation $x-y=\dfrac49p^2-\dfrac{8}{27}p^3$, \texttt{sympy} evaluates the right side to exactly $\dfrac{4}{27}$, so the candidate is $x-y=\dfrac{4}{27}$, i.e. $y=x-\dfrac{4}{27}$.

Step 6: Direct-substitution check of the singular solution

With $y=x-\dfrac{4}{27}$, $y'=1$. \texttt{sympy} confirms \[(x-y)-\left(\frac49(1)^2-\frac{8}{27}(1)^3\right)\] simplifies to exactly $0$, so $y=x-\dfrac{4}{27}$ satisfies the original equation identically. Source confirmed correct.