How to solve ODEs by converting it to Clairaut's form through suitable substitutions. Sometimes it's really confusing to find out suitable substitutions.like this one:
$(xy'-y)(x+yy')=2y'$
Which substitution should I try to put this equation in Clairaut's form?
 A: Let $v = \sqrt{y^4 + 2 y^2 x^2 + x^4 + 4 y^2 - 4 x^2 + 4}$, so that the
equation can be written as 
$$ y' = \dfrac{y^2 - x^2  + 2 \pm v}{2 x y}$$
According to Maple, an integrating factor is 
$$ {\frac {y}{ \pm \left( {x}^{2}+{y}^{2}-2 \right) v-{x}^{4}-2\,{y}^{2}{x}^{
2}-{y}^{4}+4\,{x}^{2}-4\,{y}^{2}-4}}
$$
A: Let $u=x^2+y^2$ ,
Then $\dfrac{du}{dx}=2x+2y\dfrac{dy}{dx}$
$\therefore\left(\dfrac{x}{2y}\left(\dfrac{du}{dx}-2x\right)-y\right)\dfrac{1}{2}\dfrac{du}{dx}=\dfrac{1}{y}\left(\dfrac{du}{dx}-2x\right)$
$\left(x\left(\dfrac{du}{dx}-2x\right)-2y^2\right)\dfrac{du}{dx}=4\left(\dfrac{du}{dx}-2x\right)$
$x\dfrac{du}{dx}-2x^2-2y^2=4\left(1-\dfrac{2x}{\dfrac{du}{dx}}\right)$
$x\dfrac{du}{dx}-2u=4-\dfrac{8x}{\dfrac{du}{dx}}$
$2u+4=x\dfrac{du}{dx}+\dfrac{8x}{\dfrac{du}{dx}}$
$u+2=\dfrac{x}{2}\dfrac{du}{dx}+\dfrac{4x}{\dfrac{du}{dx}}$
Let $v=x^2$ ,
Then $\dfrac{du}{dx}=\dfrac{du}{dv}\dfrac{dv}{dx}=2x\dfrac{du}{dv}$
$\therefore u+2=x^2\dfrac{du}{dv}+\dfrac{2}{\dfrac{du}{dv}}$
$u+2=v\dfrac{du}{dv}+\dfrac{2}{\dfrac{du}{dv}}$
Let $s=u+2$ ,
Then $\dfrac{ds}{dv}=\dfrac{du}{dv}$
$\therefore s=v\dfrac{ds}{dv}+\dfrac{2}{\dfrac{ds}{dv}}$
$s\dfrac{dv}{ds}=v+\dfrac{2}{\left(\dfrac{ds}{dv}\right)^2}$
$v=s\dfrac{dv}{ds}-2\left(\dfrac{dv}{ds}\right)^2$
Which reduces to Clairaut's ODE.
$\dfrac{dv}{ds}=s\dfrac{d^2v}{ds^2}+\dfrac{dv}{ds}-4\dfrac{dv}{ds}\dfrac{d^2v}{ds^2}$
$\dfrac{d^2v}{ds^2}\left(4\dfrac{dv}{ds}-s\right)=0$
$\therefore\begin{cases}\dfrac{d^2v}{ds^2}=0\\4\dfrac{dv}{ds}-s=0\end{cases}$
$\begin{cases}v=as+b\\v=\dfrac{s^2}{8}+c\end{cases}$
$\therefore\begin{cases}as+b=as-2a^2\\\dfrac{s^2}{8}+c=\dfrac{s^2}{4}-\dfrac{s^2}{8}\end{cases}$
$\begin{cases}b=-2a^2\\c=0\end{cases}$
$\therefore\begin{cases}v=as-2a^2\\v=\dfrac{s^2}{8}\end{cases}$
$\begin{cases}x^2=au+2a-2a^2\\x^2=\dfrac{(u+2)^2}{8}\end{cases}$
$\begin{cases}x^2=ax^2+ay^2+2a-2a^2\\x^2=\dfrac{(x^2+y^2+2)^2}{8}\end{cases}$
$\begin{cases}(a-1)x^2+ay^2=2a^2-2a\\x^2=\dfrac{(x^2+y^2+2)^2}{8}\end{cases}$
