\[ y=y^{\prime }x+\sqrt {4+y^{\prime 2}}\] 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 \(c\).