\[ x-yy^{\prime }=a\left ( y^{\prime }\right ) ^{2}\] Solving for \(y\) gives\begin {align} -yp & =-x+ap^{2}\nonumber \\ -y & =-\frac {x}{p}+ap\nonumber \\ y & =\frac {x}{p}-ap\tag {1}\\ y & =xf\left ( p\right ) +g\left ( p\right ) \nonumber \end {align}
Where \(f=\frac {1}{p},g=-ap\). Taking derivative and simplifying gives\begin {align*} p & =\frac {d}{dx}\left ( xf\left ( p\right ) +g\left ( p\right ) \right ) \\ & =f\left ( p\right ) +xf^{\prime }\left ( p\right ) \frac {dp}{dx}+g^{\prime }\left ( p\right ) \frac {dp}{dx} \end {align*}
But \(f\left ( p\right ) =\frac {1}{p},f^{\prime }\left ( p\right ) =\frac {-1}{p^{2}},g^{\prime }\left ( p\right ) =-a\) and the above becomes\begin {align} p & =\frac {1}{p}-\frac {x}{p^{2}}\frac {dp}{dx}-a\frac {dp}{dx}\nonumber \\ p-\frac {1}{p} & =\left ( -\frac {x}{p^{2}}-a\right ) \frac {dp}{dx} \tag {2} \end {align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=\pm 1\). Hence \(y^{\prime }=\pm 1\) or \(y=\pm x\) but these do not satisfy the ode, hence no singular solutions exist.
The general solution is when \(\frac {dp}{dx}\neq 0\) in (2). This gives the ode\begin {align*} \frac {dp}{dx} & =\frac {p-\frac {1}{p}}{-\frac {x}{p^{2}}-a}\\ & =\frac {p-p^{3}}{ap^{2}+x} \end {align*}
But this is non-linear. Hence inversion is needed. This becomes\[ \frac {dx}{dp}=\frac {-x\left ( p\right ) -ap^{2}}{p^{3}-p}\] Which is now linear in \(x\left ( p\right ) \). The solution is \begin {equation} x=\frac {-pa\sqrt {\left ( p-1\right ) \left ( p+1\right ) }\ln \left ( p+\sqrt {p^{2}-1}\right ) }{\left ( p-1\right ) \left ( p+1\right ) }+\frac {pc_{1}}{\sqrt {p-1}\sqrt {p+1}} \tag {3} \end {equation} From (1) \(y=\frac {x}{p}-ap\), hence \begin {align*} p_{1} & =\frac {1}{2}\frac {-y+\sqrt {4ax+y^{2}}}{a}\\ p_{2} & =-\frac {1}{2}\frac {y+\sqrt {4ax+y^{2}}}{a} \end {align*}
Plugging \(p_{1}\) into (3) gives one solution and Plugging \(p_{2}\) into (3) gives the second solution.