2.21 Example 21 \(y^{\prime }=2x\left ( 1-y^{2}\right ) ^{\frac {1}{2}}\)

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

Checking

\begin{align*} \frac {\partial F}{\partial y} & =\frac {x}{2}\frac {-2y}{\sqrt {1-y^{2}}}\\ & =-\frac {xy}{\sqrt {1-y^{2}}}\end{align*}

Not zero. But this is linear in \(p\). Hence no singular solution will exist using p-discriminant. Lets see. Applying p-discriminant method gives

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

The p-discriminant does not yield result since it gives \(1=0\). Lets try C-discriminant.

\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 & =\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 c-discriminant found singular solution but not p-discriminant. This is strange as books say that same \(E\) should result using both methods. Need to look more into this example.

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.

Figure 17: Singular solution envelope with general solution curves