\[ \left ( y^{\prime }\right ) ^{2}\left ( 1-y\right ) ^{2}-2+y=0 \] Applying p-discriminant method gives\begin {align*} F & =\left ( y^{\prime }\right ) ^{2}\left ( 1-y\right ) ^{2}-2+y=0\\ \frac {\partial F}{\partial y^{\prime }} & =2\left ( y^{\prime }\right ) \left ( 1-y\right ) ^{2}=0 \end {align*}
We first check that \(\frac {\partial F}{\partial y}=-2\left ( y^{\prime }\right ) ^{2}\left ( 1-y\right ) +1\neq 0\). Now we apply p-discriminant. Eliminating \(y^{\prime }\). Second equation gives \(y^{\prime }=0\) or \(\left ( 1-y\right ) ^{2}=0\). When \(y^{\prime }=0\) then first equation gives\begin {align*} -2+y & =0\\ y_{s} & =2 \end {align*}
And when \(\left ( 1-y\right ) ^{2}=0\) then \(y_{s}=1\). We check now that \(y_{s}=2\) verifies the ode. But \(y_{s}=1\) does not. Hence the only candidate \(y_{s}=2\). The primitive can be found to be \[ \Psi \left ( x,y,c\right ) =4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9\left ( x+c\right ) ^{2}=0 \] Now we have to eliminate \(c\) using the c-discriminant method. \begin {align*} \Psi \left ( x,y,c\right ) & =4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9\left ( x+c\right ) ^{2}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =-18\left ( x+c\right ) =0 \end {align*}
Second equation gives \(c=-x\). Substituting this into the first equation gives \begin {align*} 4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9\left ( x-x\right ) ^{2} & =0\\ 4\left ( 2+y\right ) \left ( y+1\right ) ^{2} & =0 \end {align*}
Hence \(y_{s}=2,y_{s}=-1\). But \(y_{s}=-1\) does not verify the ode. Hence \(y_{s}=2\). Since this is the same \(y_{s}\) obtained from p-discriminant then it is singular solution.\[ y_{s}=2 \] The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).