\[ y=xy^{\prime }+ay^{\prime }\left ( 1-y^{\prime }\right ) \] Applying p-discriminant method gives\begin {align*} F & =-y+xy^{\prime }+ay^{\prime }\left ( 1-y^{\prime }\right ) =0\\ \frac {\partial F}{\partial y^{\prime }} & =x+a\left ( 1-y^{\prime }\right ) -ay^{\prime }=0\\ & =a+x-2ay^{\prime }=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 \(y^{\prime }=\frac {a+x}{2a}\). Substituting in the first equation gives \begin {align*} -y+x\left ( \frac {a+x}{2a}\right ) +a\left ( \frac {a+x}{2a}\right ) \left ( 1-\left ( \frac {a+x}{2a}\right ) \right ) & =0\\ \frac {1}{4a}\left ( a^{2}+2ax-4ya+x^{2}\right ) & =0\\ y_{s} & =\frac {\left ( x+a\right ) ^{2}}{4a}\qquad a\neq 0 \end {align*}
Now we check if this satisfies the ode itself. We see it does. The general solution can be found to be \[ \Psi \left ( x,y,c\right ) =y-cx-ac\left ( 1-c\right ) =0 \] Hence \begin {align*} \Psi \left ( x,y,c\right ) & =y-cx-ac\left ( 1-c\right ) =0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =-x-a\left ( 1-c\right ) +ac=0 \end {align*}
Eliminating \(c\). Second equation gives \(c=\frac {a+x}{2a}\). Substituting into the general solution gives\begin {align*} y-\left ( \frac {a+x}{2a}\right ) x-a\left ( \frac {a+x}{2a}\right ) \left ( 1-\left ( \frac {a+x}{2a}\right ) \right ) & =0\\ y_{s} & =\frac {\left ( x+a\right ) ^{2}}{4a}\qquad a\neq 0 \end {align*}
Which is the same obtained using p-discriminant. Hence this is the singular solution. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\). For this, \(a=1\) was used.