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 $xy$.
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$\begingroup$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.
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