3.5.3.14 Example 14

\begin{align*} x\left ( y^{\prime }\right ) ^{2}+\left ( x-y\right ) y^{\prime }+1-y & =0\\ x\left ( y^{\prime }\right ) ^{2}+xy^{\prime }-yy^{\prime }+1-y & =0\\ y\left ( -y^{\prime }-1\right ) +x\left ( y^{\prime }\right ) ^{2}+xy^{\prime }+1 & =0 \end{align*}

Solving for \(y\)

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

Where \(f=p\) and \(g=\frac {1}{1+p}\). 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

\begin{equation} y=c_{1}x+\frac {1}{c_{1}+1} \tag {4}\end{equation}

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

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

Hence

\begin{align*} x-\frac {1}{\left ( p+1\right ) ^{2}} & =0\\ x\left ( p+1\right ) ^{2}-1 & =0\\ \left ( p+1\right ) ^{2} & =\frac {1}{x}\\ p+1 & =\pm \frac {1}{\sqrt {x}}\\ p & =\pm \frac {1}{\sqrt {x}}-1 \end{align*}

Substituting these values into (1) gives

\begin{align} y_{1} & =xp_{1}+\frac {1}{1+p_{1}}\nonumber \\ & =x\left ( \frac {1}{\sqrt {x}}-1\right ) +\frac {1}{1+\left ( \frac {1}{\sqrt {x}}-1\right ) }\nonumber \\ & =\frac {x}{\sqrt {x}}-x+\sqrt {x}\nonumber \\ & =\frac {x\sqrt {x}}{x}-x+\sqrt {x}\nonumber \\ & =2\sqrt {x}-x \tag {5}\end{align}

And substituting \(p_{2}\) into (1) gives

\begin{align} y_{1} & =xp_{1}+\frac {1}{1+p_{1}}\nonumber \\ & =x\left ( -\frac {1}{\sqrt {x}}-1\right ) +\frac {1}{1+\left ( -\frac {1}{\sqrt {x}}-1\right ) }\nonumber \\ & =-\frac {x}{\sqrt {x}}-x-\sqrt {x}\nonumber \\ & =\frac {-x\sqrt {x}}{x}-x-\sqrt {x}\nonumber \\ & =-2\sqrt {x}-x \tag {6}\end{align}

There are 3 solutions given in (4,5,6).  One is general  and two are singular.