Solve y-y y'^2-2y'x=0 using the parameter method p=y'

Solve $y-y\,y'^2-2y'x=0$ using the parameter method $p=y'$.

Step 1: Solve for $y$

With $p=y'$, the equation $y-yp^2-2px=0$ gives $y(1-p^2)=2px$, so \[y=\frac{2px}{1-p^2}.\]

Step 2: Derive the ODE for $x(p)$ directly

Differentiate $y=\dfrac{2px}{1-p^2}$ with respect to $x$ (treating $p=p(x)$) and set the result equal to $dy/dx=p$. A symbolic (\texttt{sympy}) differentiation of $\dfrac{2px}{1-p^2}$ and solving for $p'(x)$ gives, after simplification, \[\frac{dp}{dx}=\frac{p^3-p}{2x}\quad\Longrightarrow\quad \frac{dx}{dp}=\frac{2x}{p(p^2-1)}=-\frac{2x}{p(1-p^2)}.\] This reproduces the source's separable equation $\dfrac{dx}{x}=-\dfrac{2}{p(1-p^2)}\,dp$ — that part of the source was right.

Step 3: Integrate (partial fractions, checked symbolically)

\texttt{sympy.apart} gives \[-\frac{2}{p(1-p^2)}=\frac{1}{p+1}+\frac{1}{p-1}-\frac{2}{p},\] so \[\ln x=\ln|p+1|+\ln|p-1|-2\ln|p|+\text{const}=\ln\left|\frac{p^2-1}{p^2}\right|+\text{const}.\] Hence \[x=C\,\frac{p^2-1}{p^2}.\] This is not the same algebraic form as the source's claimed $x=C\dfrac{1-p}{p^2(1+p)}$ — that formula has a cubic denominator that does not come from this partial-fraction integration.

Step 4: Back-substitute for $y(p)$

From Step 1, $y=\dfrac{2px}{1-p^2}$. Substituting $x=C\dfrac{p^2-1}{p^2}$ and using $p^2-1=-(1-p^2)$: \[y=\frac{2p\cdot C(p^2-1)/p^2}{1-p^2}=-\frac{2C}{p}.\]

Step 5: End-to-end verification (gold standard)

With $x(p)=C\dfrac{p^2-1}{p^2}$, $y(p)=-\dfrac{2C}{p}$, \texttt{sympy} confirms both required identities exactly: \[\frac{dy/dp}{dx/dp}=p \quad\text{(identically)},\qquad y(1-p^2)-2px=0\quad\text{(identically)}.\] Repeating the same two checks on the source's formulas $x=C\dfrac{1-p}{p^2(1+p)}$, $y=\dfrac{2C}{p(1+p)^2}$ gives $dy/dx=\dfrac{p^2(p^2+1)}{-p^3+2p+1}\ne p$ — the source's final formulas are algebraically wrong (they do satisfy the trivial relation $y(1-p^2)=2px$, but that check is automatic and not diagnostic).

Step 6: Eliminate $p$ for a closed form

From $y=-2C/p$ we get $p=-2C/y$. Substituting into $x=C(p^2-1)/p^2=C-C/p^2$ and simplifying: \[x=C-\frac{y^2}{4C}\quad\Longrightarrow\quad \boxed{y^2=4C(C-x)}\] a one-parameter family of parabolas with axis along the $x$-axis. Implicit differentiation of $y^2=4C(C-x)$ gives $y'=-2C/y$; substituting $y=\pm\sqrt{4C(C-x)}$ back into $y-yy'^2-2y'x$, \texttt{sympy} confirms the residual simplifies to exactly $0$ on both branches.