Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$ Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$
My Attempt :
$$(y-px)^2 (1=p^2)=a^2p^2$$
$$y-px=\frac {ap}{\sqrt {1+p^2}}$$
$$y=px+\frac {ap}{\sqrt {1+p^2}}           .....(1)$$
This is Clairaut's Equation so we differentiate both sides with respect to $x$
$$\frac {dy}{dx} = p + x\cdot \frac {dp}{dx} + \frac {\sqrt {1+p^2} \cdot a \cdot \frac {dp}{dx} + ap\cdot \frac {1}{2\sqrt {1+p^2}} \cdot 2p \cdot \frac {dp}{dx}}{(1+p^2)}$$
$$p=p+x+\frac {a\sqrt {1+p^2} + \frac {ap^2}{\sqrt {1+p^2}}}{(1+p^2)} \cdot \frac {dp}{dx} $$
$$(x+\frac { a+2ap^2 }{(1+p^2)^{\frac {3}{2}}}) \cdot \frac {dp}{dx}=0$$
Either,
$$\frac {dp}{dx}=0$$
$$\textrm {so p}=c$$
Using $p=c$ in $(1)$ we get:
$$y=cx+\frac {ac}{\sqrt {1+c^2}}$$ which is the required general solution.
How do I evaluate the singular solution ?

The clairaut Eq.:
$$y=xy'+f(y')$$
D.w.r.t. $x$ to get
$$y''(x+f'(y')=0$$
Here $f'(z)$ is d.w.r. t to $z$ (the argument)
the general sdolution is gicen by $y''=0 \implies y/=C$
For the present case
$$y=xy'+\frac{ay'}{\sqrt{1+y'^2}}~~~~(1)$$
$$f'(y')=\frac{a}{(1+y'^2)^{3/2}}$$
$$\frac{x}{a}+\frac{1}{(1+y'^2)^{3/2}}=0, x<0,\implies y(x)= \pm \int \sqrt{(-a/x)^{2/3}-1} dx~~~~(2)$$
There will be no constant of integration. So the singular solution of (1) is given by (2).
