Example 1 9(y′)2 (2 − y)2 = 4(3 − y)

\begin{align*} 9\left ( y^{\prime }\right ) ^{2}\left ( 2-y\right ) ^{2} & =4\left ( 3-y\right ) \\ 9p^{2}\left ( 2-y\right ) ^{2}-4\left ( 3-y\right ) & =0 \end{align*}

Since this is quadratic in \(p\), we do not have to use elimination and can just use the quadratic discriminant

\begin{align*} b^{2}-4ac & =0\\ 0-4\left ( 9\left ( 2-y\right ) ^{2}\right ) \left ( -4\left ( 3-y\right ) \right ) & =0\\ 9\left ( 2-y\right ) ^{2}\left ( 3-y\right ) & =0\\ \left ( 2-y\right ) ^{2}\left ( 3-y\right ) & =0 \end{align*}

Comparing this to the form \(ET^{2}C=0\) we see that \(y=3\) is \(E\) (the envelop) and \(y=2\) is \(T\) (the Tac locus). This does not satisfy the ode. Let us see what happens if we use elimination method.

\begin{align*} F & =9p^{2}\left ( 2-y\right ) ^{2}-4\left ( 3-y\right ) =0\\ \frac {\partial F}{\partial p} & =18p\left ( 2-y\right ) ^{2}=0 \end{align*}

We ﬁrst check that \(\frac {\partial F}{\partial y}=-18p^{2}\left ( 2-y\right ) +4\neq 0\). Eliminating \(p\). Second equation gives \(p=0\) and \(y=2\). Substituting \(p=0\) into the ﬁrst equation gives

We now have to check if this solution satisﬁes the ode. We see it does. Hence it is the envelope. The second solution is \(y=2\) which does not satisfy the ode. This happens to be the Tac locus. We see that we obtain the same result using elimination as when using the quadratic discriminant directly.

To do the same thing use the solution, we have to ﬁrst solve the ode. The general solution (also called the primitive) can be found to be

\begin{align*} \Psi \left ( x,y,c\right ) & =\left ( x+c\right ) ^{2}-y^{2}\left ( 3-y\right ) \\ & =x^{2}+c^{2}+2xc-y^{2}\left ( 3-y\right ) \\ & =0 \end{align*}

Since this is already quadratic in \(c\), we can use the the quadratic discriminant directly. (will use \(C\) instead of \(c\) so not to confuse it with the constant of integration \(c\) in the solution)

\begin{align*} b^{2}-4aC & =0\\ \left ( 2x\right ) ^{2}-4\left ( 1\right ) \left ( -y^{2}\left ( 3-y\right ) +x^{2}\right ) & =0\\ 4x^{2}+4y^{2}\left ( 3-y\right ) -4x^{2} & =0\\ 4y^{2}\left ( 3-y\right ) & =0\\ y^{2}\left ( 3-y\right ) & =0 \end{align*}

Comparing this to the form \(EN^{2}C^{3}=0\) shows that \(E=3\) which is same as found earlier using p-discriminant and \(y=0\) is \(N\) (nodal locus).

Hence in summary, we see that \(E=3,T=2,N=0\). Only the \(E\) (envelope solution \(y=3\)) satisﬁes the ode. The others do not.

The following plot shows the singular solution as the envelope of the family of general solution plotted using diﬀerent values of \(c\) and also shows the \(N\) and \(T\) curves.

2.1 Example 1 \(9\left ( y^{\prime }\right ) ^{2}\left ( 2-y\right ) ^{2}=4\left ( 3-y\right ) \)