2.10 \(\left ( y^{\prime }\right ) ^{2}-xy^{\prime }+y=0\) \begin{align*} \left ( y^{\prime }\right ) ^{2}-xy^{\prime }+y & =0\\ p^{2}-xp+y & =0\\ F & =p^{2}-xp+y\\ & =0 \end{align*}
We first check that \(\frac {\partial F}{\partial y}=1\neq 0\) . This is quadratic in \(p\) .
\begin{align*} b^{2}-4ac & =0\\ \left ( -x\right ) ^{2}-4\left ( 1\right ) \left ( y\right ) & =0\\ x^{2}-4y & =0\\ y & =\frac {x^{2}}{4}\end{align*}
This also satisfies the ode. Hence it is \(E\) . Now we check using p-discriminant method. General
solution can be found to be
\[ \Psi \left ( x,y,c\right ) =y-xc+c^{2}=0 \]
This is quadratic in \(c\) .
\begin{align*} b^{2}-4aC & =0\\ \left ( -x\right ) ^{2}-4\left ( 1\right ) \left ( y\right ) & =0\\ x^{2}-4y & =0\\ y & =\frac {x^{2}}{4}\end{align*}
This is the same as \(y\) obtained using p-discriminant method then it is singular solution. The
following plot shows the singular solution as the envelope of the family of general solution
plotted using different values of \(c\) .