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Pengyun Li:
$y =xy'+\sqrt{(y')^2+1}$

Since we already know that $\psi(y)=\sqrt{(y')^2+1}, and \ \psi(y) = y'$

Plug $p = y'$,

we get $y = xp+\sqrt{p^2+1}$

So $\psi(p) = p,\psi'(p) = 1$

And $\psi(p) = \sqrt{p^2+1},\psi'(p) = \frac{p}{\sqrt{p^2+1}}$

Differentiate the equation w.r.t. $x$,

$pdx=pdx+(x\psi'(p)+\psi'(p))dp$

$0=(x+\frac{p}{\sqrt{p^2+1}})dp$

$dp = 0$

$\int1 dp=p=C$

Thus, $y = cx+\sqrt{c^2+1}$, is the general solution.

To get the singular solution in the parametric form,

we know that $x = -\psi'(p) = -\frac{p}{\sqrt{p^2+1}}$, hence, $\sqrt{p^2+1} =-\frac{p}{x}$,

since $y = xp+\psi(p)=xp+\sqrt{p^2+1}$, we can derive that $y = \frac{1}{\sqrt{p^2+1}} = -\frac{x}{p}$

Therefore, the singular solution in the parametric form s:
$\begin{cases}
  x = - \frac{p}{\sqrt{p^2+1}} \\
  y = - \frac{1}{p}{x}
 \end{cases}$
No it is not!!! I wrote solution in the parametric form already. Now the question is to exclude $p$ , expressing $y$ via $x$   

Pengyun Li:
$y =xy'+\sqrt{(y')^2+1}$

Since we already know that $\psi(y)=\sqrt{(y')^2+1}, and \ \psi(y) = y'$

Plug $p = y'$,

we get $y = xp+\sqrt{p^2+1}$

So $\psi(p) = p,\psi'(p) = 1$

And $\psi(p) = \sqrt{p^2+1},\psi'(p) = \frac{p}{\sqrt{p^2+1}}$

Differentiate the equation w.r.t. $x$,

$pdx=pdx+(x\psi'(p)+\psi'(p))dp$

$0=(x+\frac{p}{\sqrt{p^2+1}})dp$

$dp = 0$

$\int1 dp=p=C$

Thus, $y = cx+\sqrt{c^2+1}$, is the general solution.

To get the singular solution in the parametric form,

we know that $x = -\psi'(p) = -\frac{p}{\sqrt{p^2+1}}$, hence, $\sqrt{p^2+1} =-\frac{p}{x}$,

since $y = xp+\psi(p)=xp+\sqrt{p^2+1}$, we can derive that $y = \frac{1}{\sqrt{p^2+1}} = -\frac{x}{p}$

Therefore, the singular solution in the parametric form s:
$\begin{cases}
  x = - \frac{p}{\sqrt{p^2+1}} \\
  y = \frac{1}{\sqrt{p^2+1}}
 \end{cases}$

Since $x = -\frac{p}{\sqrt{p^2+1}}$, we can derive that $p =\pm\frac{x}{\sqrt{1-x^2}}$,

Sub into $y = -\frac{x}{p}$, we can get that:

$y = \pm\sqrt{1-x^2}$

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