\[ 1+\left ( y^{\prime }\right ) ^{2}-\frac {1}{y^{2}}=0 \] Applying p-discriminant method gives\begin {align*} F & =1+\left ( y^{\prime }\right ) ^{2}-\frac {1}{y^{2}}=0\\ \frac {\partial F}{\partial y^{\prime }} & =2y^{\prime }=0 \end {align*}
We first check that \(\frac {\partial F}{\partial y}=2\frac {1}{y^{3}}\neq 0\). Now we apply p-discriminant. Eliminating \(y^{\prime }\). Second equation gives \(y^{\prime }=0\). Hence first equation now gives \(1-\frac {1}{y^{2}}=0\) or \(y_{s}=\pm 1\). We see both solutions also satisfy the ode. The primitive can be found to be \[ \Psi \left ( x,y,c\right ) =y^{2}+\left ( x+c\right ) ^{2}-1=0 \] Now we have to eliminate \(c\) using the c-discriminant method\begin {align*} \Psi \left ( x,y,c\right ) & =y^{2}+\left ( x+c\right ) ^{2}-1=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =2\left ( x+c\right ) =0 \end {align*}
Second equation gives \(c=-x\). Substituting this into the first equation gives \begin {align*} y^{2}+\left ( x-x\right ) ^{2}-1 & =0\\ y_{s} & =\pm 1 \end {align*}
Which agrees with the p-discriminant. Hence these are the singular solutions. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).