Solve y'^3-4xyy'+8y^2=0 using the parameter method

Solve $y'^3-4xyy'+8y^2=0$ using the parameter method.

Step 1: Discard the fabricated guess

The source's proposed answer $x=2t+Ct^{1/2},\ y=4t^2+Ct^{3/2}$ was explicitly presented as an unverified guess ("I should verify by substitution") and never checked. It is discarded; the equation is re-solved from scratch below.

Step 2: Find the correct scaling (weighted homogeneity)

Try $x\to\lambda^a x$, $y\to\lambda^b y$; then $p=dy/dx\to\lambda^{\,b-a}p$. Substituting into each term of $p^3-4xyp+8y^2$: \[p^3\to\lambda^{3(b-a)},\qquad 4xyp\to\lambda^{a+b+(b-a)}=\lambda^{2b},\qquad 8y^2\to\lambda^{2b}.\] Requiring all exponents equal: $3(b-a)=2b\Rightarrow b=3a$. Taking $a=1$: $x\to\lambda x$, $y\to\lambda^3y$, $p\to\lambda^2p$ leaves the equation exactly invariant (\texttt{sympy}-confirmed: all three terms scale as $\lambda^6$).

Step 3: Exploit linearity in $x$

The equation $8y^2-4pxy+p^3=0$ is quadratic in $y$ but linear in $x$. Solving for $x$: \[x=\frac{p^3+8y^2}{4py}=\frac{p^2}{4y}+\frac{2y}{p}.\] This is the key simplification the source never found.

Step 4: Derive a separable ODE for $p(y)$

Treat $x=x(y)$ with parameter $p=p(y)$, using $\dfrac{dx}{dy}=\dfrac1p$. Differentiating $x=\dfrac{p^2}{4y}+\dfrac{2y}{p}$ symbolically in \texttt{sympy} (with $p=p(y)$) and setting the result equal to $1/p$, the equation reduces — after full simplification — to \[\frac{dp}{dy}=\frac{p}{2y},\] a simple separable ODE.

Step 5: Integrate and parametrize

$\dfrac{dp}{p}=\dfrac{dy}{2y}\Rightarrow p^2=Ky$ for a constant $K$. Writing $y=p^2/K$ and substituting into $x=\dfrac{p^2}{4y}+\dfrac{2y}{p}$ gives the clean parametric pair \[x(p)=\frac{K}{4}+\frac{2p}{K},\qquad y(p)=\frac{p^2}{K}.\]

Step 6: End-to-end verification (parametric)

\texttt{sympy} confirms, identically in $p,K$: \[\frac{dy/dp}{dx/dp}=p,\qquad\text{and}\qquad p^3-4x(p)y(p)p+8y(p)^2=0.\] Both residuals simplify to exactly $0$ — the parametric solution is genuine.

Step 7: Eliminate $p$ for a closed form

From $x=\dfrac{K}{4}+\dfrac{2p}{K}$, $p=\dfrac{K}{2}\left(x-\dfrac K4\right)$. Let $C=K/4$; substituting into $y=p^2/K$ and simplifying gives \[y=C(x-C)^2.\] Direct symbolic substitution of $y=C(x-C)^2$ into $y'^3-4xyy'+8y^2$ gives residual exactly $0$ for every real $C$ (verified with \texttt{sympy}, not just the values consistent with $K=4C\ge0$) — so this is the fully general one-parameter solution.

Step 8: $C$-discriminant test — find the singular solution

Let $G(x,y,C)=y-C(x-C)^2$. Then \[\frac{\partial G}{\partial C}=-(x-C)^2+2C(x-C)=(x-C)(3C-x)=0 \implies C=x \ \text{or}\ C=\tfrac{x}{3}.\] $C=x$ gives $G=y=0$ (the trivial $C=0$ member of the family, the $x$-axis — not a new curve). $C=x/3$ gives \[y=\frac{x}{3}\left(x-\frac x3\right)^2=\frac{4x^3}{27}.\] Substituting $y=4x^3/27$ directly into the original equation, \texttt{sympy} confirms the residual is exactly $0$: this is a genuine additional singular (envelope) solution.