#### 1.21 Example 21

$y^{\prime }=2x\left ( 1-y^{2}\right ) ^{\frac {1}{2}}$ Applying p-discriminant method gives\begin {align*} F & =y^{\prime }-2x\left ( 1-y^{2}\right ) ^{\frac {1}{2}}=0\\ \frac {\partial F}{\partial y^{\prime }} & =1=0 \end {align*}

The p-discriminant does not yield result since it gives $$1=0$$. Lets try C-discriminant. The general solution can be found to be $\Psi \left ( x,y,c\right ) =y-\sin \left ( x^{2}+2c\right ) =0$ Hence \begin {align*} \Psi \left ( x,y,c\right ) & =y-\sin \left ( x^{2}+2c\right ) =0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =-2\cos \left ( x^{2}+2c\right ) =0 \end {align*}

Hence $$x^{2}+2c=\frac {\pi }{2}$$ (there are inﬁnite solutions). Hence $$c=\frac {\pi }{4}-\frac {x^{2}}{2}$$. Substituting in the ﬁrst equation gives\begin {align*} y-\sin \left ( x^{2}+2\left ( \frac {\pi }{4}-\frac {x^{2}}{2}\right ) \right ) & =0\\ y_{s} & =\sin \left ( x^{2}+2\left ( \frac {\pi }{4}-\frac {x^{2}}{2}\right ) \right ) \\ & =\sin \left ( \frac {\pi }{2}\right ) \\ & =1 \end {align*}

And if we took $$x^{2}+2c=-\frac {\pi }{2}$$ then we now obtain $$y_{s}=-1$$. Now we check that $$y=\pm 1$$ satisfy the ode itself. We see that they do.  This is an example where

The following plot shows the singular solution as the envelope of the family of general solution plotted using diﬀerent values of $$c$$. p-discriminant does not yield result but C-discriminant does.