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?
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;