3.5.3.10 Example 10

$y=x{y}^{\prime }+ax\sqrt{1+{\left(y^{\prime }\right)}^2}$ is put in normal form (by replacing $y^{\prime }$ with $p$) which gives

$$\begin{array}{rl}\text{(1)}& y& =x(p+a\sqrt{1+p^2})& =xf\end{array}$$

where $f=p+a\sqrt{1+p^2},g=0$. Since $f\left(p\right)\ne p$ then this is d’Almbert ode. Taking derivative and simplifying gives

$$\begin{array}{rl}p& =(f+x{f}^{\prime }\frac{dp}{dx})\\ p-f& =x{f}^{\prime }\frac{dp}{dx}\end{array}$$

Using values for $f,g$ the above simplifies to

$$\begin{array}{}\text{(2A)}& -a\sqrt{1+p^2}=x(1+\frac{ap}{\sqrt{1+p^2}})\frac{dp}{dx}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ which results in $-a\sqrt{1+p^2}=0$. This gives no real solution for $p.$ Hence no singular solution exists.

The general solution is when $\frac{dp}{dx}\ne 0$ in (2A). Since (2A) is nonlinear, inversion is needed.

$$\begin{array}{rl}\frac{-a\sqrt{1+p^2}}{x+\frac{1}{2}x\frac{2ap}{\sqrt{1+p^2}}}& =\frac{dp}{dx}\\ \frac{dx}{dp}& =\frac{x(1+\frac{1}{2}\frac{2ap}{\sqrt{1+p^2}})}{-a\sqrt{1+p^2}}\\ \frac{dx}{x}& =\frac{1+\frac{1}{2}\frac{2ap}{\sqrt{1+p^2}}}{-a\sqrt{1+p^2}}dp\\ \frac{dx}{x}& =\frac{\sqrt{1+p^2}+\frac{1}{2}2ap}{-a(1+p^2)}dp\\ \frac{dx}{x}& =(-\frac{1}{a\sqrt{1+p^2}}-\frac{p}{(1+p^2)})dp\end{array}$$

Integrating gives

$$\ln x\left(p\right)=-\frac{1}{2}\ln (p^2+1)-\frac{1}{a}\mathrm{arcsinh}\left(p\right)$$

Therefore

$$\begin{array}{}\text{(3)}& x=c_1\frac{-e^{-\frac{1}{a}\left(\mathrm{arcsinh}\left(p\right)\right)}}{\sqrt{p^2+1}}\end{array}$$

There are now two choices to take. The first is by solving for $p$ from the above in terms of $x$ and substituting the result in (1) to obtain explicit solution for $y\left(x\right)$, and the second choice is by solving for $p$ algebraically from (1) and substituting the result in (3). The second choice is easier in this case but gives an implicit solution. Solving for $p$ from (1) gives

$$\begin{array}{rl}{p}_1& =-\frac{1}{x}\frac{ay+\sqrt{-a^2{x}^2+x^2+y^2}a-y}{a^2-1}\\ p_2& =\frac{1}{x}\frac{-ay+\sqrt{-a^2{x}^2+x^2+y^2}a-y}{a^2-1}\end{array}$$

Substituting each one of these solutions back in (3) gives two implicit solutions

$$\begin{array}{rl}x& =c_1\frac{-e^{-\frac{1}{a}\left(\mathrm{arcsinh}(-\frac{1}{x}\frac{ay+\sqrt{-a^2{x}^2+x^2+y^2}a-y}{a^2-1})\right)}}{\sqrt{{(-\frac{1}{x}\frac{ay+\sqrt{-a^2{x}^2+x^2+y^2}a-y}{a^2-1})}^2+1}}\\ x& =c_1\frac{-e^{-\frac{1}{a}\left(\mathrm{arcsinh}\left(\frac{1}{x}\frac{-ay+\sqrt{-a^2{x}^2+x^2+y^2}a-y}{a^2-1}\right)\right)}}{\sqrt{{\left(\frac{1}{x}\frac{-ay+\sqrt{-a^2{x}^2+x^2+y^2}a-y}{a^2-1}\right)}^2+1}}\end{array}$$