Reducing ordinary differential equation into clairaut's form Given equation is
$$x^{2}p^{2}+py\left ( 2x+y \right )+y^{2}=0$$
Where $p=\frac{\mathrm{d} y}{\mathrm{d} x}$
I know about clairaut's form and if we put $y=u$ and $xy=v$ then we will get clairaut's form. I want to learn how to guess that we need to put $y$ and $xy$.
 A: The task is to map the given equation to the Clairaut form
$$
v=qu+f(q),~~~ q=\frac{dv}{du}~\text{ or }~(v_x+v_yy')=q\,(u_x+u_yy')
$$
where $(u,v)$ is a locally bijective function of $(x,y)$ and $f$ a simple scalar function.

If there is such a Clairaut form, then the derivative of the equation factors completely, as
$$
\frac{dv}{du}=\frac{dq}{du}u+q+f'(q)\frac{dq}{du}\implies (u+f'(q))\frac{dq}{du}=0
$$
In the non-Clairaut form this may require replacing terms via the original equation.
\begin{align}
2xy'^2+2x^2y'y''+2yy'+2xy'^2&+2xyy''+2yy'^2+y^2y''+2yy'=0
\\
(2x^2y'+2xy+y^2)y''&=-2y'(2xy'+2y+yy')
\\
(2x^2y'^2+2xyy'+y^2y')y''&=-2y'^2(2xy'+2y+yy')
\\
(-2(2xyy'+y^2y'+y^2)+2xyy'+y^2y')y''&=-2y'^2(2xy'+2y+yy')
\\
(2y^2+2xyy'+y^2y')y''&=2y'^2(2xy'+2y+yy')
\\
(2xy'+2y+yy')(yy''-2y'^2)&=0
\end{align}
which means that solutions have segments where $(4xy+y^2)'=0$ or $y'=Cy^2$.
The "linear" solution family is then $\frac1y=ax+b$ via the second factor, while the first factor, after used to eliminate $y'$ from the original equation, gives the singular curve or envelope.

More precisely, substituting back $p=y'=Cy^2$ with $y\ne 0$ into the original equation gives
$$
x^2C^2y^4+Cy^3(2x+y)+y^2=0
\\
(Cxy)^2+2(Cxy)+Cy^2+1=0
\\
(1+Cxy)^2+Cy^2=0
$$
This only has solutions for $C=-b^2\le0$ where then
$$
\frac1y=b^2x\pm b
$$
which has the indicated form with $a=b^2$. The solution families for both signs and $b \in\Bbb R$ coincide, so only keep the plus sign.

This gets the form of a standard example of the class of Clairaut DE after dividing by $x$,
$$
\frac1{xy}=b\frac1x+b^2
$$
so that $u=\frac1x$, $v=\frac1{xy}=\frac{u}{y(1/u)}$, $f(q)=q^2$. Now
$$
q=\frac{d}{du}\frac{u}{y(1/u)}=\frac1{y(1/u)}+\frac{y'(1/u)}{uy(1/u)^2}
=\frac{y+xy'(x)}{y(x)^2}
$$
so that the Clairaut form (for $y\ne 0$) transforms to
$$
\frac1{xy}=\frac{y+xp}{xy^2}+\frac{y^2+2xyp+x^2p^2}{y^4}
\\
0=py^2+y^2+2xyp+x^2p^2
$$
which is indeed the given equation.
