Singular solution of DE problem Find the singular solution of the given de
$$y=x{dy\over dx}+a{dy\over dx}\left[1+\left({dy\over dx}\right)^2\right]^{-1\over 2}$$
My attempt : this is a clairaut's form of DE of the form $y=xp+ap(1+p^2)^{-1\over 2}$
so its solution is $y=xc+ac(1+c^2)^{-1\over 2}$.
Now we need to eliminate $p$ between $y(x)$ and $y'(x)$
going in this way and solving, differentiating i reach a point where i get
$$-x={a\over ({1+p^2})^{3 \over 2}} $$ then just reduce for $p$ and putting in the initial equation for y, thus getting a very complex equation. does it look good?
 A: Why not? In fact you have $$x=\frac{-a}{\sqrt{(1+c^2)^3}}\longrightarrow\sqrt{1+c^2}=\sqrt[3]{\frac{-a}{x}},~~~c=\pm\sqrt{\left(\frac{a}{x}\right)^{2/3}-1}$$ and so you will have, while we take the $+$ of above; $$y_{\text{singilar}}(x)=x\sqrt{(-a/x)^{2/3}-1}+\frac{a\sqrt{(-a/x)^{2/3}-1}}{(-a/x)^{1/3}}$$ For example and plotting the envelope here, I set $a=14$ and then;

A: You get your singular solution from $$-x={a\over ({1+p^2})^{3 \over 2}}$$
Process for finding Singular solution Solution: $$-x={a\over ({1+p^2})^{3 \over 2}}$$
$$\implies ({1+p^2})^{3 \over 2}=-\frac{a}{x}$$
$$\implies 1+p^2=(\frac{a}{x})^{2 \over 3}$$
$$\implies p^2=\frac{a^{2 \over 3}-x^{2 \over 3}}{x^{2 \over 3}}$$
$$\implies p=\frac{dy}{dx}=\sqrt{\frac{a^{2 \over 3}-x^{2 \over 3}}{x^{2 \over 3}}}$$
$$dy=\sqrt{\frac{a^{2 \over 3}-x^{2 \over 3}}{x^{2 \over 3}}} ~dx\qquad . . . . . . .(1)$$
Putting $~a^{2 \over 3}-x^{2 \over 3}=u\implies du=-\frac{2}{3}x^{-{1 \over 3}}~dx$
From $(1)$, $$dy=-\frac{3}{2}u^{\frac{1}{2}}du$$Integrating,  $$y=-u^{\frac{3}{2}}=-(a^{2 \over 3}-x^{2 \over 3})^{\frac{3}{2}}$$
$$\implies x^{2 \over 3}+y^{2 \over 3}=a^{2 \over 3}$$This is the required singular solution. 
