Solve (y')^3+y^2=xy y' using the parameter method

Solve $(y')^3+y^2=xy\,y'$ using the parameter method.

Step 1: Why the source's answer is discarded

With $p=y'$, the source correctly reaches $x=\dfrac{p^2}{y}+\dfrac{y}{p}$, then abandons a direct approach ("this still has $p$ on the right side... this is getting complicated"), explicitly shows that a guessed ansatz $p=Ct^2$ fails, and finally asserts without derivation that $x=t+C/t^2,\ y=Ct+C^2/t^3$ is "the standard result," with the claimed verification never carried out. Substituting this candidate into $p^3+y^2-xyp$ with \texttt{sympy} (using $p=\frac{dy/dt}{dx/dt}$) gives a residual that is not identically zero. The source's final answer is therefore confirmed fabricated and is discarded; the problem is rederived from scratch below.

Step 2: Reduce via $t=y/p$

Substituting $p=y/t$ into $p^3+y^2=xyp$:

Dividing by $y^2$ ($y\neq0$) and multiplying by $t^3$ gives the algebraic relation

This turns the problem into a Lagrange-type equation in the new parameter $t$, together with the compatibility relation $p=y/t=xt-t^2$.

Step 3: Derive the ODE for $t(x)$ (sympy)

Treating $t$ as a function of $x$ in $y=xt^2-t^3$, \texttt{sympy} computes $dy/dx$ by the chain rule and sets it equal to $p=y/t=xt-t^2$ (the two expressions for $p$ must agree). Solving the resulting equation for $t'(x)$ gives, exactly,

i.e. $(2x-3t)\,dt=(x-2t)\,dx$, confirming the structure the source began to set up (this matches $t'\cdot(2x-3t)=x-2t$).

Step 4: Solve the homogeneous ODE

The right-hand side is homogeneous of degree $0$ in $(x,t)$. Calling \texttt{sympy.dsolve} directly on $t'(x)=\dfrac{x-2t}{2x-3t}$ returns, in closed form,

Equivalently, squaring $(3t-2x)^2=x^2+C_1$ gives the implicit algebraic invariant $3t^2-4xt+x^2=K$ (with $K=C_1/3$), which inverted for $x$ in terms of $t$ gives $x=2t\pm\sqrt{t^2+K}$ — the same relation, with the parameter roles of $x$ and $t$ swapped, confirming internal consistency.

Step 5: Assemble the parametric solution and reconstruct $y$

Writing $s=\sqrt{t^2+C}$ and using $x=2t+\sigma s$ (either sign $\sigma=\pm1$) together with $y=xt^2-t^3=t^2(x-t)=t^2(t+\sigma s)$ gives the parametric solution, with parameter $t$ and arbitrary constant $C$:

Step 6: Gold-standard verification

\texttt{sympy} computes $dx/dt$ and $dy/dt$ for this parametric family (both signs $\sigma=\pm1$) and forms the ratio $\dfrac{dy/dt}{dx/dt}$; it simplifies identically to $p=t(x-t)=t^2+\sigma st$, confirming $dy/dx=p$ exactly. Substituting $p=dy/dt \big/ dx/dt$, $x(t)$, $y(t)$ directly into $p^3+y^2-xyp$ and simplifying with \texttt{sympy} returns $0$ identically for both branches — a complete, independent confirmation of the derived solution.