#### 1.14 Example 14

$1+\left ( y^{\prime }\right ) ^{2}-\frac {1}{y^{2}}=0$ 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 $$c$$.