#### 1.8 Example 8

$xy^{\prime }+y^{\prime }-\left ( y^{\prime }\right ) ^{2}-y=0$ Applying p-discriminant method gives\begin {align*} F & =xy^{\prime }+y^{\prime }-\left ( y^{\prime }\right ) ^{2}-y=0\\ \frac {\partial F}{\partial y^{\prime }} & =x+1-2y^{\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 $$y^{\prime }=\frac {x+1}{2}$$. Substituting in the ﬁrst equation gives\begin {align*} x\left ( \frac {x+1}{2}\right ) +\frac {x+1}{2}-\left ( \frac {x+1}{2}\right ) ^{2}-y & =0\\ y_{s} & =\frac {1}{4}x^{2}+\frac {1}{2}x+\frac {1}{4} \end {align*}

Now we verify this satisﬁes the ode. We see it does. Now we have to ﬁnd the general solution. It will be$y_{c}=c+cx-c^{2}$ Now we have to eliminate $$c$$ using the c-discriminant method\begin {align*} \Psi \left ( x,y,c\right ) & =-y+c+cx-c^{2}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =1+x-2c=0 \end {align*}

Second equation gives $$1+x-2c=0$$ or $$c=\frac {1+x}{2}$$. Substituting into the ﬁrst gives\begin {align*} -y+\left ( \frac {1+x}{2}\right ) +\left ( \frac {1+x}{2}\right ) x-\left ( \frac {1+x}{2}\right ) ^{2} & =0\\ \frac {1}{4}x^{2}+\frac {1}{2}x-y+\frac {1}{4} & =0\\ y_{s} & =\frac {1}{4}x^{2}+\frac {1}{2}x+\frac {1}{4} \end {align*}

We see this is the same as $$y_{s}$$ from the p-discriminant method. Hence it is a singular solution. The following plot shows the singular solution as the envelope of the family of general solution plotted using diﬀerent values of $$c$$.