1.2.3 \(27y-8p^{3}=0\)

\begin{align*} 27y-8\left ( y^{\prime }\right ) ^{3} & =0\\ 27y-8p^{3} & =0 \end{align*}

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

\begin{align*} F & =27y-8p^{3}=0\\ \frac {\partial F}{\partial p} & =-24p^{2}=0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}=27\neq 0\).  Second equation gives \(p=0\). Substituting this into first equation gives \(27y=0\) or \(y=0\). We see this also satisfies the ode. Hence it is \(E\) (the envelope). The general solution can be found as

\[ \Psi \left ( x,y,c\right ) =y^{2}-\left ( x+c\right ) ^{3}=0 \]

Applying c-discriminant

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

Second equation gives \(\left ( x+c\right ) ^{2}=0\) or \(c=-x\). From first equation this gives \(y^{2}=0\) or \(y=0\). This is the same as \(y_{s}\) found from p-discriminant, hence

\[ y=0 \]

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

Figure 4: Singular solution envelope with general solution curves