\(y=x\left ( y^{\prime }\right ) ^{2}+\left ( y^{\prime }\right ) ^{2}\) is put in normal form (by replacing \(y^{\prime }\) with \(p\)) which gives\begin {align} y & =xp^{2}+p^{2}\tag {1}\\ & =xf+g\nonumber \end {align}
where \(f=p^{2},g=p^{2}\). Taking derivative and simplifying gives\begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) +\left ( g^{\prime }\frac {dp}{dx}\right ) \\ p & =f+\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx}\\ p-f & =\left ( xf^{\prime }+g^{\prime }\right ) \frac {dp}{dx} \end {align*}
Using values for \(f,g\) the above simplifies to\begin {equation} p-p^{2}=\left ( 2xp+2p\right ) \frac {dp}{dx} \tag {2A} \end {equation} The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(p=0\) or \(p=1\). Substituting these values in (1) gives the singular solutions\begin {align} y_{1}\left ( x\right ) & =0\tag {3}\\ y_{2}\left ( x\right ) & =x+1 \tag {4} \end {align}
The general solution is found by finding \(p\) from (2A). Since (2A) is not linear in \(p\), then inversion is needed. Writing (A2) as\[ \frac {p\left ( 1-p\right ) }{2p\left ( x+1\right ) }=\frac {dp}{dx}\] Inverting gives\begin {align*} \frac {dx}{dp} & =\frac {2\left ( x+1\right ) }{\left ( 1-p\right ) }\\ \frac {dx}{dp}-x\frac {2}{\left ( 1-p\right ) } & =\frac {2}{\left ( 1-p\right ) } \end {align*}
This is now linear \(x\left ( p\right ) \). The solution is\[ x=\frac {C^{2}}{\left ( p-1\right ) ^{2}}-1 \] Solving for \(p\) gives\begin {align*} \frac {C^{2}}{\left ( p-1\right ) ^{2}} & =x+1\\ \left ( p-1\right ) ^{2} & =\frac {C^{2}}{x+1}\\ \left ( p-1\right ) & =\pm \frac {C}{\sqrt {x+1}}\\ p & =1\pm \frac {C}{\sqrt {x+1}} \end {align*}
Substituting the above in (1) gives the general solutions\[ y=\left ( x+1\right ) p^{2}\] Therefore\begin {align*} y\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac {C}{\sqrt {x+1}}\right ) ^{2}\\ y\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac {C}{\sqrt {x+1}}\right ) ^{2} \end {align*}
The solution \(y_{1}\left ( x\right ) =0\) found earlier can not be obtained from the above general solution hence it is singular solution. But \(y_{2}\left ( x\right ) =x+1\) can be obtained from the general solution when \(C=0\). Hence there are only three solutions, they are\begin {align*} y_{1}\left ( x\right ) & =0\\ y_{2}\left ( x\right ) & =\left ( x+1\right ) \left ( 1+\frac {C}{\sqrt {x+1}}\right ) ^{2}\\ y_{3}\left ( x\right ) & =\left ( x+1\right ) \left ( 1-\frac {C}{\sqrt {x+1}}\right ) ^{2} \end {align*}