Solve x√(1+y'^2)=y' using the parameter method

Solve $x\sqrt{1+y'^2}=y'$ using the parameter method.

Step 1: Solve for $x$

With $p=y'$: $x\sqrt{1+p^2}=p\Rightarrow x=\dfrac{p}{\sqrt{1+p^2}}$.

Step 2: Differentiate (checked symbolically)

\texttt{sympy} gives \[\frac{dx}{dp}=\frac{d}{dp}\left[\frac{p}{\sqrt{1+p^2}}\right]=\frac{1}{(1+p^2)^{3/2}},\] matching the source exactly.

Step 3: Integrate for $y$

$dy=p\,dx=\dfrac{p}{(1+p^2)^{3/2}}\,dp$. \texttt{sympy.integrate} confirms \[y=-\frac{1}{\sqrt{1+p^2}}+C,\] matching the source exactly.

Step 4: Eliminate $p$

From $x=\dfrac{p}{\sqrt{1+p^2}}$, $x^2=\dfrac{p^2}{1+p^2}\Rightarrow p^2=\dfrac{-x^2}{x^2-1}$. Substituting into $y=C-\dfrac{1}{\sqrt{1+p^2}}$ and simplifying with \texttt{sympy}, the expression \[x^2+(y-C)^2-1\] simplifies to exactly $0$, confirming \[x^2+(y-C)^2=1.\]

Step 5: Verify by implicit differentiation, with branch care

Differentiating $x^2+(y-C)^2=1$ gives $y'=\dfrac{x}{C-y}$. Substituting the two branches $y=C\pm\sqrt{1-x^2}$: the lower branch $y=C-\sqrt{1-x^2}$ satisfies $x\sqrt{1+y'^2}-y'\equiv 0$ exactly (checked numerically and symbolically at several points). The upper branch instead satisfies $x\sqrt{1+y'^2}=-y'$, i.e. it solves a sign-flipped equation. This is expected: as $p$ ranges over all reals in Steps 1–3, $y=C-1/\sqrt{1+p^2}$ never exceeds $C$, so the construction only ever traces the lower semicircle. The implicit relation $x^2+(y-C)^2=1$ is the correct algebraic envelope of the construction, understood with $y\le C$ (equivalently $y-C=-\sqrt{1-x^2}$).