3.5.3.13 Example 13

\begin{align} y & =\frac {xy^{\prime }+x\left ( y^{\prime }\right ) ^{2}-\left ( y^{\prime }\right ) ^{2}}{y^{\prime }+1}\nonumber \\ & =\frac {xp+xp^{2}-p^{2}}{p+1}\nonumber \\ & =xp-\frac {p^{2}}{p+1}\tag {1}\\ & =xf+g\nonumber \end{align}

Where \(f=p\) and \(g=-\frac {p^{2}}{p+1}\). Since \(f\left ( p\right ) =p\) then this is Clairaut ode. Taking derivative of the above w.r.t. \(x\) gives

\begin{align*} p & =\frac {d}{dx}\left ( xp+g\left ( p\right ) \right ) \\ p & =p+\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\\ 0 & =\left ( x+g^{\prime }\left ( p\right ) \right ) \frac {dp}{dx}\end{align*}

The general solution is given by

\begin{align*} \frac {dp}{dx} & =0\\ p & =c_{1}\end{align*}

Substituting this in (1) gives the general solution

\[ y=xc_{1}-\frac {c_{1}^{2}}{c_{1}+1}\]

The term \(\left ( x+g^{\prime }\left ( p\right ) \right ) =0\) is used to find singular solutions.

\begin{align*} x+g^{\prime }\left ( p\right ) & =x+\frac {d}{dp}\frac {1}{p}\\ & =x-\frac {1}{p^{2}}\end{align*}

Hence \(x-\frac {1}{p^{2}}=0\) or \(p=\pm \frac {1}{\sqrt {x}}\). Substituting these back in (1) gives

\begin{align} y_{1}\left ( x\right ) & =xp+\frac {1}{p}\nonumber \\ & =x\frac {1}{\sqrt {x}}+\sqrt {x}\nonumber \\ & =2\sqrt {x}\tag {3}\\ y_{2}\left ( x\right ) & =-x\sqrt {\frac {1}{x}}-\sqrt {x}\nonumber \\ & =-2\sqrt {x} \tag {4}\end{align}

Eq. (2) is the general solution and (3,4) are the singular solutions.