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 \(c\).