3.5.3.14 Example 14

$$\begin{array}{rl}x{\left(y^{\prime }\right)}^2+(x-y)y^{\prime }+1-y& =0\\ x{\left(y^{\prime }\right)}^2+x{y}^{\prime }-y{y}^{\prime }+1-y& =0\\ y(-y^{\prime }-1)+x{\left(y^{\prime }\right)}^2+x{y}^{\prime }+1& =0\end{array}$$

Solving for $y$

$$\begin{array}{rl}y& =\frac{-x{\left(y^{\prime }\right)}^2-x{y}^{\prime }-1}{-y^{\prime }-1}\\ & =\frac{-x{p}^2-xp-1}{-p-1}\\ & =\frac{x{p}^2+xp+1}{p+1}\\ & =x\left(\frac{p^2+p}{p+1}\right)+\frac{1}{1+p}\\ & =xp+\frac{1}{1+p}\\ \text{(1)}& & =xf+g\end{array}$$

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

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

The general solution is given by

$$\begin{array}{rl}\frac{dp}{dx}& =0\\ p& =c_1\end{array}$$

Substituting this in (1) gives the general solution

$$\begin{array}{}\text{(4)}& y=c_{1}x+\frac{1}{c_1+1}\end{array}$$

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

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

Hence

$$\begin{array}{rl}x-\frac{1}{{(p+1)}^2}& =0\\ x{(p+1)}^2-1& =0\\ {(p+1)}^2& =\frac{1}{x}\\ p+1& =\pm \frac{1}{\sqrt{x}}\\ p& =\pm \frac{1}{\sqrt{x}}-1\end{array}$$

Substituting these values into (1) gives

$$\begin{array}{rl}{y}_1& =x{p}_1+\frac{1}{1+p_1}\\ & =x(\frac{1}{\sqrt{x}}-1)+\frac{1}{1+(\frac{1}{\sqrt{x}}-1)}\\ & =\frac{x}{\sqrt{x}}-x+\sqrt{x}\\ & =\frac{x\sqrt{x}}{x}-x+\sqrt{x}\\ \text{(5)}& & =2\sqrt{x}-x\end{array}$$

And substituting $p_2$ into (1) gives

$$\begin{array}{rl}{y}_1& =x{p}_1+\frac{1}{1+p_1}\\ & =x(-\frac{1}{\sqrt{x}}-1)+\frac{1}{1+(-\frac{1}{\sqrt{x}}-1)}\\ & =-\frac{x}{\sqrt{x}}-x-\sqrt{x}\\ & =\frac{-x\sqrt{x}}{x}-x-\sqrt{x}\\ \text{(6)}& & =-2\sqrt{x}-x\end{array}$$

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