2.8 Example 8 \(xy^{\prime }+y^{\prime }-\left ( y^{\prime }\right ) ^{2}-y=0\)

\begin{align*} xy^{\prime }+y^{\prime }-\left ( y^{\prime }\right ) ^{2}-y & =0\\ xp+p-p^{2}-y & =0\\ p^{2}-p\left ( 1+x\right ) +y & =0 \end{align*}

This is quadratic in \(p\).

\begin{align*} b^{2}-4ac & =0\\ \left ( -\left ( 1+x\right ) \right ) ^{2}-4\left ( 1\right ) \left ( y\right ) & =0\\ \left ( 1+x\right ) ^{2}-4y & =0\\ y & =\frac {\left ( 1+x\right ) ^{2}}{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

\begin{align*} y & =c+cx-c^{2}\\ 0 & =y-c\left ( 1+x\right ) +c^{2}\end{align*}

This is quadratic in \(c\).

\begin{align*} b^{2}-4aC & =0\\ \left ( -\left ( 1+x\right ) \right ) ^{2}-4\left ( 1\right ) \left ( y\right ) & =0\\ \left ( 1+x\right ) ^{2}-4y & =0\\ y & =\frac {\left ( 1+x\right ) ^{2}}{4}\end{align*}

We see this is the same as \(y\) 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 \(c\).

Figure 5: Singular solution envelope with general solution curves