2.17 \(\left ( y-xy^{\prime }\right ) ^{2}-\left ( y^{\prime }\right ) ^{2}=1\) \begin{align*} \left ( y-xy^{\prime }\right ) ^{2}-\left ( y^{\prime }\right ) ^{2} & =1\\ \left ( y-xp\right ) ^{2}-p^{2} & =1\\ F & =1+p^{2}-\left ( y-xp\right ) ^{2}\\ & =1+p^{2}-\left ( y^{2}+x^{2}p^{2}-2yxp\right ) \\ & =1+p^{2}-y^{2}-x^{2}p^{2}+2yxp\\ & =p^{2}\left ( 1-x^{2}\right ) +p\left ( 2yx\right ) +1-y^{2}\\ & =0 \end{align*}
We first check that \(\frac {\partial F}{\partial y}=2\left ( y-xp\right ) \neq 0\) . This is quadratic in \(p\) . Hence
\begin{align*} b^{2}-4ac & =0\\ \left ( 2yx\right ) ^{2}-4\left ( 1-x^{2}\right ) \left ( 1-y^{2}\right ) & =0\\ 4x^{2}+4y^{2}-4 & =0\\ x^{2}+y^{2}-1 & =0\\ y & =\pm \sqrt {1-x^{2}}\end{align*}
Both of these solutions verify the ode. Hence they are both singular solutions. The primitive
can be found to be
\begin{align*} \Psi \left ( x,y,c\right ) & =y-xc\pm \sqrt {1+c^{2}}\\ & =0 \end{align*}
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
\[ y=\pm \sqrt {1-x^{2}}\]
The following plot shows the singular solution
as the envelope of the family of general solution plotted using different values of
\(c\) .