#### 1.10 Example 10

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