Solve y=y'x+√(1+y'^2)

Solve $y=y'x+\sqrt{1+y'^2}$.

Step 1: Identify Clairaut form and general solution

$f(p)=\sqrt{1+p^2}$. For any constant $c$, $y=cx+\sqrt{1+c^2}$, $y'\equiv c$. \texttt{sympy} confirms \[y-\left(xy'+\sqrt{1+y'^2}\right)\equiv0\] — the line family is genuine.

Step 2: Singular-solution condition

$f'(p)=\dfrac{p}{\sqrt{1+p^2}}$, so $x+f'(p)=0$ gives $x+\dfrac{p}{\sqrt{1+p^2}}=0$.

Step 3: Solve parametrically and identify the correct branch

Rather than inverting for $p(x)$ directly (which introduces spurious sign branches), parametrize by $p$ itself: \[x(p)=-\frac{p}{\sqrt{1+p^2}},\qquad y(p)=x(p)\,p+\sqrt{1+p^2}.\] \texttt{sympy.simplify} reduces $y(p)$ to exactly $\dfrac{1}{\sqrt{1+p^2}}$, and confirms $x(p)^2+y(p)^2-1\equiv0$ for all real $p$ — the singular locus lies on the unit circle.

Step 4: Only the upper semicircle is traced

Since $y(p)=1/\sqrt{1+p^2}>0$ for \emph{every} real $p$, this construction only ever reaches points with $y>0$: the \textbf{upper} unit semicircle. As $x^2=p^2/(1+p^2)=1-y^2$ with $y>0$, this is $y=\sqrt{1-x^2}$, $-1

Step 5: Confirm the source's sign choice for $p(x)$

Differentiating $y=\sqrt{1-x^2}$ gives $p=y'=\dfrac{-x}{\sqrt{1-x^2}}$, exactly the branch the source selected (\texttt{sympy}: $p_{\text{computed}}-p_{\text{source}}\equiv0$).

Step 6: Direct-substitution verification

With $y=\sqrt{1-x^2}$, $p=-x/\sqrt{1-x^2}$, \texttt{sympy} (checked both symbolically after rationalizing, and numerically at several sample points in $(-1,1)$) confirms \[y-\left(xp+\sqrt{1+p^2}\right)\equiv0.\] The singular solution and its branch/sign are confirmed correct.