A differential equation (nonlinear First-Order) how to solve this equation:
$(Px-y)(Py+x)=h^2P$
that $P=\frac{dy}{dx}$
and $h$ is a constant.
 A: Assume $h\neq0$ for the key case:
Let $u=x^2+y^2$ ,
Then $\dfrac{du}{dx}=2x+2y\dfrac{dy}{dx}$
$\therefore\left(\dfrac{x}{2y}\left(\dfrac{du}{dx}-2x\right)-y\right)\dfrac{1}{2}\dfrac{du}{dx}=\dfrac{h^2}{2y}\left(\dfrac{du}{dx}-2x\right)$
$\left(x\left(\dfrac{du}{dx}-2x\right)-2y^2\right)\dfrac{du}{dx}=2h^2\left(\dfrac{du}{dx}-2x\right)$
$x\dfrac{du}{dx}-2x^2-2y^2=2h^2\left(1-\dfrac{2x}{\dfrac{du}{dx}}\right)$
$x\dfrac{du}{dx}-2u=2h^2-\dfrac{4h^2x}{\dfrac{du}{dx}}$
$2u+2h^2=x\dfrac{du}{dx}+\dfrac{4h^2x}{\dfrac{du}{dx}}$
$u+h^2=\dfrac{x}{2}\dfrac{du}{dx}+\dfrac{2h^2x}{\dfrac{du}{dx}}$
Let $v=x^2$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=2x\dfrac{du}{dv}$
$\therefore u+h^2=x^2\dfrac{du}{dv}+\dfrac{h^2}{\dfrac{du}{dv}}$
$u+h^2=v\dfrac{du}{dv}+\dfrac{h^2}{\dfrac{du}{dv}}$
Let $s=u+h^2$ ,
Then $\dfrac{ds}{dv}=\dfrac{du}{dv}$
$\therefore s=v\dfrac{ds}{dv}+\dfrac{h^2}{\dfrac{ds}{dv}}$
$s\dfrac{dv}{ds}=v+\dfrac{h^2}{\left(\dfrac{ds}{dv}\right)^2}$
$v=s\dfrac{dv}{ds}-h^2\left(\dfrac{dv}{ds}\right)^2$
Which reduces to Clairaut's ODE.
$\dfrac{dv}{ds}=s\dfrac{d^2v}{ds^2}+\dfrac{dv}{ds}-2h^2\dfrac{dv}{ds}\dfrac{d^2v}{ds^2}$
$\dfrac{d^2v}{ds^2}\left(2h^2\dfrac{dv}{ds}-s\right)=0$
$\therefore\begin{cases}\dfrac{d^2v}{ds^2}=0\\2h^2\dfrac{dv}{ds}-s=0\end{cases}$
$\begin{cases}v=as+b\\v=\dfrac{s^2}{4h^2}+c\end{cases}$
$\therefore\begin{cases}as+b=as-h^2a^2\\\dfrac{s^2}{4h^2}+c=\dfrac{s^2}{2h^2}-\dfrac{s^2}{4h^2}\end{cases}$
$\begin{cases}b=-h^2a^2\\c=0\end{cases}$
$\therefore\begin{cases}v=as-h^2a^2\\v=\dfrac{s^2}{4h^2}\end{cases}$
$\begin{cases}x^2=au+h^2a-h^2a^2\\x^2=\dfrac{(u+h^2)^2}{4h^2}\end{cases}$
$\begin{cases}x^2=ax^2+ay^2+h^2a-h^2a^2\\x^2=\dfrac{(x^2+y^2+h^2)^2}{4h^2}\end{cases}$
$\begin{cases}(a-1)x^2+ay^2=h^2a^2-h^2a\\x^2=\dfrac{(x^2+y^2+h^2)^2}{4h^2}\end{cases}$
