\[ x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x=0 \] Solving for \(y\) gives\begin {align} y & =x\left ( \frac {1}{2}y^{\prime }+2\frac {1}{y^{\prime }}\right ) \tag {1}\\ & =x\left ( \frac {1}{2}p+2\frac {1}{p}\right ) \nonumber \\ y & =xf\nonumber \end {align}
where \(f=\frac {1}{2}p+2\frac {1}{p},g=0\). Taking derivative and simplifying gives\begin {align*} p & =\left ( f+xf^{\prime }\frac {dp}{dx}\right ) \\ p-f & =xf^{\prime }\frac {dp}{dx} \end {align*}
Using values for \(f,g\) the above simplifies to\begin {align} p-\frac {1}{2}p-2\frac {1}{p} & =x\left ( \frac {1}{2}-\frac {2}{p^{2}}\right ) \frac {dp}{dx}\nonumber \\ \frac {1}{2}p-\frac {2}{p} & =x\left ( \frac {1}{2}-\frac {2}{p^{2}}\right ) \frac {dp}{dx} \tag {2A} \end {align}
The singular solution is found by setting \(\frac {dp}{dx}=0\) which results in \(\frac {1}{2}p-\frac {2}{p}=0\) or \(\frac {1}{2}p^{2}-2=0\) or \(p^{2}=4\) or \(p=\pm 2\). Hence \(y=\pm 2x\) are the singular solutions.
The general solution is when \(\frac {dp}{dx}\neq 0\) in (2A). Since (2A) is nonlinear, inversion is needed. General solution can be shown to be\begin {equation} y=-\frac {1}{2}\left ( -\frac {x^{2}}{c_{1}^{2}}-4\right ) c_{1} \tag {3} \end {equation} Will now show a more general method to find singular solution that works for any first order ode. This requires finding the general solution above first. Let the general solution be
\begin {align*} \Phi \left ( x,y,c\right ) & =0\\ & =y+\frac {1}{2}\left ( -\frac {x^{2}}{c_{1}^{2}}-4\right ) c_{1} \end {align*}
The ode is \begin {align*} F\left ( x,y,y^{\prime }\right ) & =0\\ & =x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x \end {align*}
First we find the p-discriminant curve. This is found by eliminating \(y^{\prime }\) from \begin {align*} F & =0\\ \frac {\partial F}{\partial y^{\prime }} & =0 \end {align*}
Or\begin {align*} x\left ( y^{\prime }\right ) ^{2}-2yy^{\prime }+4x & =0\\ 2xy^{\prime }-2y & =0 \end {align*}
Second equation gives \(y^{\prime }=\frac {y}{x}\). Substituting into first equation gives \(x\left ( \frac {y}{x}\right ) ^{2}-2y\left ( \frac {y}{x}\right ) +4x=0\) or \(\frac {y^{2}}{x}-2\frac {y^{2}}{x}+4x=0\) or \(y=\pm 2x\). These are the candidate singular solutions\[ y_{s}=\pm 2x \] Next, we verify these satisfy the ode itself. We see both do. Next we have to check that for an arbitrary point \(x_{0}\) the following two equations are satisfied\begin {align*} y_{g}\left ( x_{0}\right ) & =y_{s}\left ( x_{0}\right ) \\ y_{g}^{\prime }\left ( x_{0}\right ) & =y_{s}^{\prime }\left ( x_{0}\right ) \end {align*}
Where \(y_{g}\left ( x\right ) \) is the general solution obtained above in (3). Starting with \(y_{s}=2x\) the above two equations now become\begin {align*} -\frac {1}{2}\left ( -\frac {x_{0}^{2}}{c_{1}^{2}}-4\right ) c_{1} & =2x_{0}\\ -\frac {1}{2}\left ( -\frac {2x_{0}}{c_{1}^{2}}\right ) c_{1} & =2 \end {align*}
Or\begin {align*} \frac {x_{0}^{2}}{2c_{1}}+2c_{1} & =2x_{0}\\ \frac {x_{0}}{c_{1}} & =2 \end {align*}
Second equation gives \(c_{1}=\frac {x_{0}}{2}\). Using this in first equation gives\begin {align*} \frac {x_{0}^{2}}{2\frac {x_{0}}{2}}+2\left ( \frac {x_{0}}{2}\right ) & =2x_{0}\\ x_{0}+x_{0} & =2x_{0}\\ 2x_{0} & =2x_{0} \end {align*}
Which shows it is satisfied. Hence this shows that \(y_{s}=2x\) is indeed a singular solution. Now we have to do the same for second \(y_{s}=-2x\). Hence the steps of this method are the following