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 ?
Best Answer
Both sign variants of the square roots give solutions, you can include this by using also the solutions where $a$ is replaced by $-a$.
In $y=px+f(p)$ the derived equation is $0=y''(x+f'(y'))$ so that any solution will be composed of segments where one of the factors is zero. You already found the lines $y'=p=c$.
linear solution family for $|a|=2$, blue lines for $a=2$ and red lines for $a=-2$
The other factor gives with $$ f'(p)=\frac{a}{\sqrt{1+p^2}}-\frac{ap^2}{\sqrt{1+p^2}^3}=\frac{a}{\sqrt{1+p^2}^3} $$ the equation $$ \sqrt{1+p^2}=-\sqrt[3]{\frac{a}{x}}\implies p=\pm\sqrt{\left(\frac{a}{x}\right)^{2/3}-1},~~ax<0 $$ which can be inserted into the original equation \begin{align} y(x)&=p\left(x-a\sqrt[3]{\frac{x}{a}}\right) \\ &=\pm\sqrt{\left(\frac{a}{x}\right)^{2/3}-1}\left(x-a^{2/3}x^{1/3}\right) \end{align} This last formula does not depend on the sign of $a$. Switching the parts to have a unique formula above and below the $x$ axis results in $$ y(x)=\pm \left(a^{2/3}-x^{2/3}\right)^{3/2},~~|x|\le|a|. $$