\[ \left ( y^{\prime }\right ) ^{2}-xy^{\prime }+y=0 \] Applying p-discriminant method gives\begin {align*} F & =\left ( y^{\prime }\right ) ^{2}-xy^{\prime }+y=0\\ \frac {\partial F}{\partial y^{\prime }} & =2y^{\prime }-x=0 \end {align*}
We first check that \(\frac {\partial F}{\partial y}=1\neq 0\). Now we apply p-discriminant. Eliminating \(y^{\prime }\). Second equation gives \(y^{\prime }=\frac {1}{2}x\). Substituting into the first equation gives\begin {align*} \left ( \frac {1}{2}x\right ) ^{2}-x\left ( \frac {1}{2}x\right ) +y & =0\\ y_{s} & =\frac {1}{4}x^{2} \end {align*}
This also satisfies the ode. Now we check using p-discriminant method. General solution can be found to be\[ \Psi \left ( x,y,c\right ) =xc-c^{2}=0 \] Now we have to eliminate \(c\) using the c-discriminant method\begin {align*} \Psi \left ( x,y,c\right ) & =y-xc+c^{2}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =-x+2c=0 \end {align*}
Second equation gives \(c=\frac {1}{2}x\). First equation gives \begin {align*} y-x\left ( \frac {1}{2}x\right ) +\left ( \frac {1}{2}x\right ) ^{2} & =0\\ y_{s} & =\frac {1}{4}x^{2} \end {align*}
Since this is the same as \(y_{s}\) 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 \(c\).