Solve x=yy'+a(y')^2 using the parameter method (roles of x…

Solve $x=yy'+a(y')^2$ using the parameter method (roles of $x$ and $y$ swapped relative to standard Clairaut form), and determine the true singular solutions.

Step 1: Why the source's linear ODE is wrong

Let $p=y'$. Differentiating $x=yp+ap^2$ with \texttt{sympy} and imposing $dx/dx=1$ gives $1=p^2+p'(y+2ap)$, so $p'=\dfrac{1-p^2}{y+2ap}$ (this part matches the source). But then $\dfrac{dy}{dp}=\dfrac{dy/dx}{dp/dx}=\dfrac{p}{p'}=\dfrac{p(y+2ap)}{1-p^2}$ — the source's claimed $\dfrac{dy}{dp}=\dfrac{y+2ap}{1-p^2}$ is missing the factor of $p$ (that expression is actually $dx/dp$, obtained from a different, valid identity $dx/dp=\partial_p(yp+ap^2)/(1-p\,\partial_y(yp+ap^2))$ — but the source uses it as if it were $dy/dp$, which is inconsistent). \texttt{sympy}'s gold-standard check applied to the source's claimed solution $y=-2a+C/\sqrt{1-p^2}$ confirms this: $\dfrac{dy/dp}{dx/dp}-p$ does not simplify to $0$. The source's general solution is wrong and is discarded.

Step 2: Solve the corrected linear ODE (sympy)

The correct equation is $\dfrac{dy}{dp}-\dfrac{p}{1-p^2}\,y=\dfrac{2ap^2}{1-p^2}$. \texttt{sympy.dsolve} returns (for $p^2>1$, and after substituting $\ln(p+\sqrt{p^2-1})=\operatorname{arccosh}p$):

and $x(p)=yp+ap^2$. \texttt{sympy} confirms this satisfies the ODE identically (residual $0$), and the gold-standard check $\dfrac{dy/dp}{dx/dp}-p$ also simplifies to $0$ — a fully independent confirmation that this (considerably more involved) formula, not the source's, is the true general solution.

Step 3: Both of the source's singular-solution candidates are wrong

The source offers two mutually inconsistent candidates: $y=-2a$ (called "the envelope," from an unjustified step) and $x^2=4a^2(y+2a)$ (from eliminating $p$ using an unmotivated relation $p=-y/(2a)$). Testing $y=-2a$ (constant, so $y'=0$): the original equation requires $x=y\cdot0+a\cdot0=0$ identically, which fails for $x\ne0$. Testing $x=-y^2/(4a)$ (equivalent to the source's second form): implicit differentiation gives $y'=-2a/y$, and \texttt{sympy} substitution into $x-(yy'+a(y')^2)$ gives a nonzero residual $-\tfrac{y^2}{4a}+2a-\tfrac{4a^3}{y^2}$. Both candidates fail.

Step 4: The true singular solutions (Lagrange fixed-point test, solved-for-$x$ form)

For an equation solved for $x$, $x=\varphi(p)y+\psi(p)$ with $\varphi(p)=p,\ \psi(p)=ap^2$, the genuine singular solutions occur where $1-p\varphi(p)=0$ (the solved-for-$x$ analogue of the Lagrange fixed-point condition), i.e. $1-p^2=0\Rightarrow p=\pm1$ — precisely the points excluded from Step 2's solution. Substituting $p=1$: $x=y+a$, i.e. $y=x-a$; substituting $p=-1$: $x=-y+a$, i.e. $y=a-x$. \texttt{sympy} confirms both are genuine solutions: with $y'\equiv1$, $yy'+a(y')^2=y+a=x$ identically; with $y'\equiv-1$, $yy'+a(y')^2=-y+a=x$ identically.