#### 1.16 Example 16

$\left ( y^{\prime }\right ) ^{2}\left ( 1-y\right ) ^{2}-2+y=0$ 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 $$c$$.