3.5.3.13 Example 13

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

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

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

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

$$y=x{c}_1-\frac{c_1^2}{c_1+1}$$

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

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

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

$$\begin{array}{rl}{y}_1\left(x\right)& =xp+\frac{1}{p}\\ & =x\frac{1}{\sqrt{x}}+\sqrt{x}\\ \text{(3)}& & =2\sqrt{x}{y}_2\left(x\right)& =-x\sqrt{\frac{1}{x}}-\sqrt{x}\\ \text{(4)}& & =-2\sqrt{x}\end{array}$$

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