2.16 Example 16 \(\left ( y^{\prime }\right ) ^{2}\left ( 1-y\right ) ^{2}-2+y=0\)

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

Hence

\begin{align} F & =p^{2}\left ( 1-y\right ) ^{2}-2+y\tag {1}\\ & =0\nonumber \end{align}

We first check that \(\frac {\partial F}{\partial y}=-2\left ( y^{\prime }\right ) ^{2}\left ( 1-y\right ) +1\neq 0\). Since the ode is quadratic in \(p\), we can use the more direct method which is the discriminant of the quadratic equation instead of the elimination method. The discriminant of (1) is

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

Comparing this to \(ET^{2}C=0\) shows that \(y=2\) is \(E\), i.e. the envelope singular solution which also satisfies the ode, while \(y=1\) is \(T\) which is Tac locus. This does not satisfy the ode but it plotted below to show geometrically what it means.

The primitive can be found to be

\[ \Psi \left ( x,y,c\right ) =4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9\left ( x+c\right ) ^{2}=0 \]

Since this is also quadratic in \(c\), we can use the more direct method which is the discriminant of the quadratic equation instead of the elimination method. Hence

\begin{align*} 4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9\left ( x^{2}+c^{2}+2cx\right ) & =0\\ 4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9x^{2}-9c^{2}-18cx & =0 \end{align*}

Therefore

\begin{align*} b^{2}-4aC & =0\\ \left ( -18x\right ) ^{2}-4\left ( -9\right ) \left ( 4\left ( 2-y\right ) \left ( y+1\right ) ^{2}-9x^{2}\right ) & =0\\ \left ( -18x\right ) ^{2}+144\left ( 2-y\right ) \left ( y+1\right ) ^{2}-324x^{2} & =0\\ 144\left ( 2-y\right ) \left ( y+1\right ) ^{2} & =0\\ \left ( 2-y\right ) \left ( y+1\right ) ^{2} & =0 \end{align*}

Comparing to general form \(EN^{2}C^{3}=0\) shows that \(y=2\) is \(E\) which agrees with what was found using p-discriminant as expected, and \(y=-1\) is \(N\) which is nodal locus \(N\) since it shows to power of two. Notice that \(N\) can only show up using c-discriminant method.

The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\), and also show the nodal locus \(N\) shown as solid green line and \(T\) (Tac locus) drawn as solid blue line.  

The above result of \(E,N,T\) can also be found using elimination method. But since the equation are already quadratic in \(p,c\), it can be easier to just use the quadratic equation discriminant directly. As metioned before, elimination method is more general since it works for any equation and not just quadratic.

Figure 13: Singular solution envelope with general solution curves