#### 1.9 Example 9

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

We ﬁrst check that $$\frac {\partial F}{\partial y}=1\neq 0$$.  Now we apply p-discriminant.  Second equation gives $$x\sqrt {4+y^{\prime 2}}+y^{\prime }=0$$ which gives$y^{\prime }=\pm \frac {2x}{\sqrt {1-x^{2}}}$ Trying the negative root and substituting it in $$F=0$$ gives\begin {align*} y-\left ( -\frac {2x}{\sqrt {1-x^{2}}}\right ) x-\sqrt {4+\left ( -\frac {2x}{\sqrt {1-x^{2}}}\right ) ^{2}} & =0\\ y+\frac {2x^{2}}{\sqrt {1-x^{2}}}-\sqrt {4+\frac {4x^{2}}{1-x^{2}}} & =0\\ y+\frac {2x^{2}}{\sqrt {1-x^{2}}}-2\frac {\sqrt {1-x^{2}+x^{2}}}{\sqrt {1-x^{2}}} & =0\\ y+\frac {2x^{2}}{\sqrt {1-x^{2}}}-\frac {2}{\sqrt {1-x^{2}}} & =0\\ y+\frac {2\left ( x^{2}-1\right ) }{\sqrt {1-x^{2}}} & =0\\ y+\frac {2\left ( x^{2}-1\right ) \sqrt {1-x^{2}}}{1-x^{2}} & =0\\ y+\frac {-2\left ( 1-x^{2}\right ) \sqrt {1-x^{2}}}{1-x^{2}} & =0\\ y-2\sqrt {1-x^{2}} & =0\\ y_{s} & =2\sqrt {1-x^{2}} \end {align*}

Which satisﬁes the ode.  The general solution can be found to be $\Psi \left ( x,y,c\right ) =y-xc-\sqrt {4+c^{2}}=0$ Now we have to eliminate $$c$$ using the c-discriminant method\begin {align*} \Psi \left ( x,y,c\right ) & =y-xc-\sqrt {4+c^{2}}=0\\ \frac {\partial \Psi \left ( x,y,c\right ) }{\partial c} & =-x-\frac {2c}{2\sqrt {4+c^{2}}}=0 \end {align*}

Second equation gives $c=\pm \frac {2x}{\sqrt {1-x^{2}}}$ Taking the negative root, and substituting into the ﬁrst equation gives\begin {align*} y-x\left ( \frac {-2x}{\sqrt {1-x^{2}}}\right ) -\sqrt {4+\left ( \frac {2x}{\sqrt {1-x^{2}}}\right ) ^{2}} & =0\\ y+\left ( \frac {2x^{2}}{\sqrt {1-x^{2}}}\right ) -2\sqrt {1+\frac {x^{2}}{1-x^{2}}} & =0\\ y+\left ( \frac {2x^{2}}{\sqrt {1-x^{2}}}\right ) -\frac {2}{\sqrt {1-x^{2}}} & =0\\ y-\frac {2\left ( 1-x^{2}\right ) }{\sqrt {1-x^{2}}} & =0\\ y-\frac {2\left ( 1-x^{2}\right ) \sqrt {1-x^{2}}}{1-x^{2}} & =0\\ y-2\sqrt {1-x^{2}} & =0\\ y_{s} & =2\sqrt {1-x^{2}} \end {align*}

Which is the same obtained using the p-discriminant. Hence$y_{s}=2\sqrt {1-x^{2}}$ Is singular solution. We have to try the other root also. But graphically, the above seems to be the only valid singular solution.  The following plot shows the singular solution as the envelope of the family of general solution plotted using diﬀerent values of $$c$$.