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