2.12 Example 12 \(\left ( y^{\prime }\right ) ^{2}x+y^{\prime }y\ln y-y^{2}\left ( \ln y\right ) ^{4}=0\)

\begin{align*} \left ( y^{\prime }\right ) ^{2}x+y^{\prime }y\ln y-y^{2}\left ( \ln y\right ) ^{4} & =0\\ p^{2}x+py\ln y-y^{2}\left ( \ln y\right ) ^{4} & =0 \end{align*}

Applying p-discriminant method gives

\begin{align*} F & =p^{2}x+py\ln y-y^{2}\left ( \ln y\right ) ^{4}=0\\ \frac {\partial F}{\partial p} & =2px+y\ln y=0 \end{align*}

We first check that \(\frac {\partial F}{\partial y}\neq 0\).  Now we apply p-discriminant. Eliminating \(p\). Second equation gives \(p=-\frac {y}{2x}\ln y\). Substituting into the first equation gives

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

Hence we obtain the solutions

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

Or

\begin{align} y & =0\tag {1}\\ y & =1\tag {2}\\ 1+4x\left ( \ln y\right ) ^{2} & =0 \tag {3}\end{align}

The solution \(y=0\) does not satisfy the ode. But \(y=1\) does. The solution \(1+4x\left ( \ln y\right ) ^{2}=0\) gives \(y=\left \{ \begin {array} [c]{c}e^{\frac {-i}{2\sqrt {x}}}\\ e^{\frac {i}{2\sqrt {x}}}\end {array} \right . \) But these do not satisfy the ode.

The primitive can be found to be

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

Now we have to eliminate \(c\) using the c-discriminant method

\begin{align*} \Psi \left ( x,y,c\right ) & =y-e^{\frac {c}{c^{2}-x}}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =\left ( \frac {1}{c^{2}-x}-\frac {2c^{2}}{\left ( c^{2}-x\right ) ^{2}}\right ) e^{\frac {c}{c^{2}-x}}=0 \end{align*}

Second equation gives \(\frac {1}{c^{2}-x}-\frac {2c^{2}}{\left ( c^{2}-x\right ) ^{2}}=0\) or \(e^{\frac {c}{c^{2}-x}}=0\). For the first one, \(c=\pm \sqrt {-x}\). Substituting \(\sqrt {-x}\) in first equation gives

\begin{align} y-e^{\frac {\sqrt {-x}}{-x-x}} & =0\nonumber \\ y & =e^{\frac {\sqrt {-x}}{-2x}}\nonumber \\ \ln y & =\frac {\sqrt {-x}}{-2x}\nonumber \\ \left ( \ln y\right ) ^{2} & =\frac {-x}{4x^{2}}\nonumber \\ 4x\left ( \ln y\right ) ^{2}+1 & =0\nonumber \\ y_{s} & =\left \{ \begin {array} [c]{c}e^{\frac {-i}{2\sqrt {x}}}\\ e^{\frac {i}{2\sqrt {x}}}\end {array} \right . \tag {4}\end{align}

These do not satisfy the ode. Can not obtain \(y=1\) solution using c-discriminant.

See paper by C.N. SRINIVASINGAR, example 4. The following plot shows the singular solution above as the envelope of the family of general solution plotted using different values of \(c\). Added also \(y_{s}=1\).

Figure 9: Singular solution envelope with general solution curves