Solve y=xy'+ax√(1+(y')^2) and determine whether a singular…

Solve $y=xy'+ax\sqrt{1+(y')^2}$ and determine whether a singular solution exists.

Step 1: The source's Clairaut claim and its asserted answer are both wrong

Standard Clairaut form requires $y=xp+f(p)$ with $f$ depending on $p$ alone; here the extra term is $ax\sqrt{1+p^2}$, which contains $x$ explicitly, so this equation is not Clairaut. The source's own attempt to differentiate it gets "complex," and it then simply asserts, for constant $p=c$, that $y=x\big(c+a\sqrt{1+c^2}\big)$ is the general solution, with no verification. But if $y=mx$ with $m=c+a\sqrt{1+c^2}$ is to be a genuine solution, its own derivative $y'=m$ must equal the $p=c$ used to build it, i.e. $c+a\sqrt{1+c^2}=c$. \texttt{sympy.solve} on this equation for $a$ returns only $a=0$. So for $a\ne0$, the source's asserted general solution is false — confirmed exactly as suspected.

Step 2: Correct reformulation — $y=x\cdot g(p)$

Rewrite the equation as $y=x\big(p+a\sqrt{1+p^2}\big)=x\,g(p)$ with $g(p)=p+a\sqrt{1+p^2}$, a genuinely homogeneous-type relation (right side is $x$ times a function of $p$ alone). Differentiating $y=x\,g(p)$ with \texttt{sympy}, using $dy/dx=p$: $p=g(p)+x\,g'(p)\,p'$. Since $g(p)-p=a\sqrt{1+p^2}$, this rearranges to $x\,g'(p)\,p'=-a\sqrt{1+p^2}$, i.e. the separable equation

Step 3: Integrate exactly with sympy

\texttt{sympy.integrate} applied to $-\dfrac{g'(p)}{a\sqrt{1+p^2}}=-\dfrac{1}{a\sqrt{1+p^2}}-\dfrac{p}{1+p^2}$ returns, exactly,

Exponentiating (and using $\operatorname{arcsinh}p=\ln\!\big(p+\sqrt{1+p^2}\big)$, verified equal by \texttt{sympy} both symbolically and numerically) gives the parametric general solution

Step 4: Gold-standard and direct verification (sympy)

\texttt{sympy} computes $dx/dp$ and $dy/dp$ for this parametric family and forms $\dfrac{dy/dp}{dx/dp}-p$: it simplifies identically to $0$, confirming $dy/dx=p$ exactly. Direct substitution of $x(p),y(p)$ into $y-\big(xp+ax\sqrt{1+p^2}\big)$ also returns $0$ identically (as it must, since $y=x\,g(p)$ by construction) — a full, independent confirmation of the true general solution.

Step 5: No singular solution exists (proved, not assumed)

The source's Clairaut-style envelope trick is invalid here (Step 1), so it is discarded rather than reused. Instead: \texttt{sympy} proves generally that no line $y=mx$ (constant slope $m$) can solve the equation for $a\ne0$ — substituting $y=mx,\ y'=m$ into $y-\big(xy'+ax\sqrt{1+(y')^2}\big)$ gives the residual $-ax\sqrt{1+m^2}$, which is nonzero for any $a\ne0,\ x\ne0$. Separately, the naive $p$-discriminant of $F=y-x\,g(p)$ gives $\partial F/\partial p=-x\,g'(p)=0$, i.e. $g'(p)=0$; \texttt{sympy.solve} finds real roots $p_0=\pm1/\sqrt{a^2-1}$ only when $|a|>1$, and substituting the resulting candidate line $y=x\,g(p_0)$ back into the original equation gives a nonzero residual in every case — exactly the same spurious-line failure proved generally above. Since both routes rule out any constant-slope candidate, and the family's own tangent direction (slope $p$) is never parallel to the "radial" direction of the solution family (slope $g(p)$) unless $g(p)=p$, which forces $a=0$, this equation has no singular solution for any $a\ne0$.