Many general and singular solutions from Clairaut DE The ODE $$ (y-xy')^2-y'^2=1$$
I think since it is first order second degree ODE , it can be solved by writing it in this form y=F(y',x) and then differentiating wrt x , or writing it in the form x=F(y,y') and then differentiating wrt y . (please tell me if this is wrong).
Instead of doing this, I solved it in the following way by transforming it to two Clairaut DEs:
I obtained two general solutions and four singular solutions, I feel that some solutions are refused but which solutions and why : 
First solution :
$$y=xy'+\sqrt{1+y'^2}$$
Differentiate wrt x
$$y''(x+\frac{y'}{\sqrt{1+y'^2}})=0$$
so the general solution is
$$y=xc+\sqrt{1+c^2}$$
The singular solution is when $$x=-\frac{y'}{\sqrt{1+y'^2}}$$
or we can write this as $$y'=\pm \frac{x}{\sqrt{1-x^2}}$$
Eliminate y' from the DE
$$y=xy'+\sqrt{1+y'^2}$$
So we have 2 singular solutions since y' is positive or negative , the singular solutions are:
$$y=\frac{x^2+1}{\sqrt{1-x^2}}$$
$$y=\frac{-x^2+1}{\sqrt{1-x^2}}=\sqrt{1-x^2}$$
The second solution , we will take the negative square root for y 
$$y=xy'-\sqrt{1+y'^2}$$
Apply same steps ,so differentiate wrt x
$$y''(x-\frac{y'}{\sqrt{1+y'^2}})=0$$
so the second general solution is
$$y=xc-\sqrt{1+c^2}$$
The singular solution is when $$x=\frac{y'}{\sqrt{1+y'^2}}$$
or we can write this as $$y'=\pm \frac{x}{\sqrt{1-x^2}}$$
Eliminate y' from the DE
$$y=xy'-\sqrt{1+y'^2}$$
So the two other singular solutions will be
$$y=\frac{x^2-1}{\sqrt{1-x^2}}=-\sqrt{1-x^2}$$
$$y=\frac{-x^2-1}{\sqrt{1-x^2}}$$
 A: You found the linear solutions
$$y=cx+s\sqrt{1+c^2}\tag{LINSOL}$$
in both cases $s=\pm 1$  for the constant sign of the square root in
$$y=y'x+s\sqrt{1+y'^2}\tag{ODE}$$ 
correctly. However, for the singular solutions one has to observe that in the equation
$$
x+s\frac{y'}{\sqrt{1+y'^2}}=0\tag1
$$
the signs of $x$ and $y'$ are coupled. Thus while $x^2=\frac{y'^2}{1+y'^2}$ implies indeed
$$
|y'|=\frac{|x|}{\sqrt{1-x^2}},\tag2
$$
there is only one solution for each value of $s$,
$$
y'=-s\frac{x}{\sqrt{1-x^2}},\tag3
$$
Inserting back into the original equation (ODE) gives
$$
y=-s\frac{x^2}{\sqrt{1-x^2}}+s\frac1{\sqrt{1-x^2}}=s\sqrt{1-x^2}\tag{SINGSOL}
$$
A: The given differential equation is $$(y-xy')^2-y'^2=1$$
$$\implies y=xy'\pm\sqrt{1+y'^2}$$ which is of Clairaut form.
Differentiating both side with respect to $x$, we have
$$y'=y'+xy''\pm\frac{y'y''}{\sqrt{1+y'^2}}\implies y''(x\pm\frac{y'}{\sqrt{1+y'^2}})=0$$which gives $y''=0\implies y'=c,\quad$ where $c$ is an arbitrary constant
Hence the general solution of the given differential solution is $$y=cx\pm\sqrt{1+c^2}$$ where $c$ is an arbitrary constant.
Also $$x\pm\frac{y'}{\sqrt{1+y'^2}}=0\implies y'=\pm\frac{x}{\sqrt{1-x^2}}$$Integrating, $$y=\pm\sqrt{1-x^2}$$These are the singular solutions.
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Conclusion: 


*

*When we consider the given differential equation  $$(y-xy')^2-y'^2=1$$ is of the form $$ y=xy'+\sqrt{1+y'^2}$$then the general solution of the differential equation is $$y=cx+\sqrt{1+c^2}$$ where $c$ is an arbitrary constant $\qquad$ and the singular solution be $$y=\pm\sqrt{1-x^2}.$$

*Again when we consider the given differential equation  $$(y-xy')^2-y'^2=1$$ is of the form $$ y=xy'-\sqrt{1+y'^2}$$then the general solution of the differential equation is $$y=cx-\sqrt{1+c^2}$$ where $c$ is an arbitrary constant $\qquad$ and the singular solution be $$y=\pm\sqrt{1-x^2}.$$
