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