2.2 Example 2 \(\left ( y^{\prime }\right ) ^{2}=xy\)

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

This has form \(ET^{2}C=0\). We see that \(y=0\) satisfies the ode, hence it is \(E\) (because \(C\) do not satisfy the ode). We found no \(T\) nor \(C\).

To apply c-discriminant we have to find the general solution. It will be

\begin{align*} y_{1} & =\frac {1}{36}\left ( 4x^{3}-12x^{\frac {3}{2}}c+9c^{2}\right ) \\ y_{2} & =\frac {1}{36}\left ( 4x^{3}+12x^{\frac {3}{2}}c+9c^{2}\right ) \end{align*}

Hence we have two general solutions. These can be written as

\begin{align*} \Psi _{1}\left ( x,y,c\right ) & =y-\frac {1}{36}\left ( 4x^{3}-12x^{\frac {3}{2}}c+9c^{2}\right ) =0\\ & =y+\frac {1}{3}cx^{\frac {3}{2}}-\frac {1}{4}c^{2}-\frac {1}{9}x^{3}=0\\ \Psi _{2}\left ( x,y,c\right ) & =y-\frac {1}{36}\left ( 4x^{3}+12x^{\frac {3}{2}}c+9c^{2}\right ) =0\\ & =y-\frac {1}{3}cx^{\frac {3}{2}}-\frac {1}{4}c^{2}-\frac {1}{9}x^{3}=0 \end{align*}

Since these are quadratic in \(c\), we can use the quadratic discriminant. First equation above gives

\begin{align*} b^{2}-4aC & =0\\ \left ( \frac {1}{3}x^{\frac {3}{2}}\right ) ^{2}-4\left ( -\frac {1}{4}\right ) \left ( y-\frac {1}{9}x^{3}\right ) & =0\\ \frac {1}{9}x^{3}+y-\frac {1}{9}x^{3} & =0\\ y & =0 \end{align*}

Comparing to \(EN^{2}C^{3}=0\) shows this is \(E\). Same as before. Second solution will give same result.

Hence \(y=0\) is the singular solution. No \(T,N,C\,\) were found. The following plot shows the singular solution as the envelope of the family of general solution plotted using different values of \(c\).

Figure 3: Singular solution envelope with general solution curves