#### 1.12 Example 12

$\left ( y^{\prime }\right ) ^{2}x+y^{\prime }y\ln y-y^{2}\left ( \ln y\right ) ^{4}=0$ Applying p-discriminant method gives\begin {align*} F & =\left ( y^{\prime }\right ) ^{2}x+y^{\prime }y\ln y-y^{2}\left ( \ln y\right ) ^{4}=0\\ \frac {\partial F}{\partial y^{\prime }} & =2y^{\prime }x+y\ln y=0 \end {align*}

We ﬁrst check that $$\frac {\partial F}{\partial y}\neq 0$$.  Now we apply p-discriminant. Eliminating $$y^{\prime }$$. Second equation gives $$y^{\prime }=-\frac {y}{2x}\ln y$$. 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 & =\infty \\ \frac {1}{4x}-\frac {1}{2x}-\left ( \ln y\right ) ^{2} & =0 \end {align*}

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

The solution $$y=0$$ does not satisfy the ode. Same for $$y=\infty$$.  The solution $$1+4x\left ( \ln y\right ) ^{2}=0$$ gives $$y=\left \{ \begin {array} [c]{c}e^{\frac {-i}{2\sqrt {x}}}\\ e^{\frac {i}{2\sqrt {x}}}\end {array} \right .$$ and only the ﬁrst one satisﬁes the ode and for negative $$x$$ only. 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 $$\frac {1}{c^{2}-x}-\frac {2c^{2}}{\left ( c^{2}-x\right ) ^{2}}=0$$. Hence $$c=\pm \sqrt {-x}$$. 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}

Substituting $$-\sqrt {-x}$$ in ﬁrst equation gives same solution. But only the ﬁrst one satisﬁes the ode for $$x<0$$. This is the same as obtained using p-discriminant, hence it is a singular solution.

There is another singular solution $$y_{s}=1$$. But now I do not know how to ﬁnd this. I need to look more into this. 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 $$y_{s}=1$$.