2.14 Example 14 \(1+\left ( y^{\prime }\right ) ^{2}-\frac {1}{y^{2}}=0\)

\begin{align*} 1+\left ( y^{\prime }\right ) ^{2}-\frac {1}{y^{2}} & =0\\ F & =1+p^{2}-\frac {1}{y^{2}}\\ & =0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=2\frac {1}{y^{3}}\neq 0\). This is quadratic in \(p\).

\begin{align*} b^{2}-4ac & =0\\ \left ( 0\right ) -4\left ( 1\right ) \left ( 1-\frac {1}{y^{2}}\right ) & =0\\ 1-\frac {1}{y^{2}} & =0\\ y^{2} & =1\\ y & =\pm 1 \end{align*}

We see both solutions also satisfy the ode. The primitive can be found to be

\begin{align*} \Psi \left ( x,y,c\right ) & =y^{2}+\left ( x+c\right ) ^{2}-1\\ & =y^{2}+x^{2}+c^{2}+2xc-1\\ & =0 \end{align*}

This is quadratic in \(c\).

\begin{align*} b^{2}-4aC & =0\\ \left ( 2x\right ) ^{2}-4\left ( 1\right ) \left ( y^{2}+x^{2}-1\right ) & =0\\ 4x^{2}-4y^{2}-4x^{2}+4 & =0\\ y^{2} & =1\\ y & =\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 \(c\).

Figure 11: Singular solution envelope with general solution curves