\[ \left ( y-xy^{\prime }\right ) ^{2}-\left ( y^{\prime }\right ) ^{2}=1 \] Applying p-discriminant method gives\begin {align*} F & =\left ( y-xy^{\prime }\right ) ^{2}-\left ( y^{\prime }\right ) ^{2}-1=0\\ \frac {\partial F}{\partial y^{\prime }} & =2\left ( y-xy^{\prime }\right ) \left ( -x\right ) -2y^{\prime }=0\\ & =-2yx+2x^{2}y^{\prime }-2y^{\prime }=0 \end {align*}
We first check that \(\frac {\partial F}{\partial y}=2\left ( y-xy^{\prime }\right ) \neq 0\). Now we apply p-discriminant. Eliminating \(y^{\prime }\). Second equation gives \begin {align*} -2yx+\left ( 2x^{2}-2\right ) y^{\prime } & =0\\ y^{\prime } & =\frac {2yx}{\left ( 2x^{2}-2\right ) } \end {align*}
Substituting into \(F=0\) gives\begin {align*} \left ( y-x\left ( \frac {2yx}{\left ( 2x^{2}-2\right ) }\right ) \right ) ^{2}-\left ( \frac {2yx}{\left ( 2x^{2}-2\right ) }\right ) ^{2}-1 & =0\\ -\frac {1}{x^{2}-1}\left ( x^{2}+y^{2}-1\right ) & =0\\ x^{2}+y^{2}-1 & =0\\ y^{2} & =1-x^{2}\\ y_{s} & =\pm \sqrt {1-x^{2}} \end {align*}
Both of these solution verify the ode. The primitive can be found to be \[ \Psi \left ( x,y,c\right ) =y-xc\pm \sqrt {1+c^{2}}=0 \] Eliminating \(c\). First solution gives\begin {align*} \Psi _{1}\left ( x,y,c\right ) & =y-xc+\sqrt {1+c^{2}}=0\\ \frac {\partial \Psi _{1}\left ( x,y,c\right ) }{\partial c} & =-x+\frac {1}{2}\frac {2c}{\sqrt {1+c^{2}}}=0 \end {align*}
Second equation gives \(-x+\frac {1}{2}\frac {2c}{\sqrt {1+c^{2}}}=0\) or \(c=x\sqrt {\frac {1}{1-x^{2}}}\). Substituting into the first equation above gives\begin {align*} y-x\left ( x\sqrt {\frac {1}{1-x^{2}}}\right ) +\sqrt {1+\left ( x\sqrt {\frac {1}{1-x^{2}}}\right ) ^{2}} & =0\\ y-x^{2}\sqrt {\frac {1}{1-x^{2}}}+\sqrt {1+\frac {x^{2}}{1-x^{2}}} & =0\\ y-x^{2}\sqrt {\frac {1}{1-x^{2}}}+\sqrt {\frac {1-x^{2}+x^{2}}{1-x^{2}}} & =0\\ y-x^{2}\sqrt {\frac {1}{1-x^{2}}}+\sqrt {\frac {1}{1-x^{2}}} & =0\\ y+\sqrt {\frac {1}{1-x^{2}}}\left ( 1-x^{2}\right ) & =0\\ y & =\sqrt {\frac {1}{1-x^{2}}}\left ( x^{2}-1\right ) \\ & =\frac {\left ( x^{2}-1\right ) }{\sqrt {1-x^{2}}}\\ & =\frac {\left ( x^{2}-1\right ) \sqrt {1-x^{2}}}{1-x^{2}}\\ & =-\sqrt {1-x^{2}} \end {align*}
Which is given by p-discriminant above. Hence it is singular solution. If we try \(\Psi _{2}\left ( x,y,c\right ) =y-xc-\sqrt {1+c^{2}}=0\) we also can verify the second singular solution. Hence\[ y_{s}=\pm \sqrt {1-x^{2}}\] The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).