I have a few differential equations with their substitution to form Clairaut form, but is there any standard method to get the intuition of the substitution?
examples:
- $y = 2xp + y^2 p^3$, for transforming , put $y^2 = Y$
- $y + px = x^4 p^2$, for transforming , put $1/x = X$
- $y = 3px + 6 y^2 p^2$, for transforming , put $y^3 = Y$
and few other transformation cases.
Note : $p = \frac{dy}{dx}$
I understand that reducing to Clairaut's form involves suitable substitution so as to bring it in the form of $Y = P X + f(P)$ but i am unable to form any intuition about what such substitutions might be , as the above equations seem complicated with more than one combination of variables and $p$.
Help appreciated.
There is a structure that all equations that are equivalent to a Clairaut form have to satisfy. Taking the derivative of the equation modulo the original equation has to factor into one term containing the second derivative and one without it.
Take the second equation, $y+xy'=x^4y'^2$. Its derivative is $$ 2y'+xy''=4x^3y'^2+2x^4y'y''\iff (xy''+2y')(1-2x^3y')=0 $$ so that we get a direct factorization. The first term gives the family $y'=-Cx^{-2}$, $y=Cx^{-1}+D$ which tells us that the transformation $X=x^{-1}$ produces the solution family as family of lines.
In the third equation $y = 3xy' + 6 y^2 y'^2$ the derivative is $$ y'=3y'+3xy''+12yy'^3+12y^2y'y''\iff0=3y''(x+4y^2y')+2y'(1+6yy'^2) $$ Now the original equation can be rearranged as $y(1+6yy'^2)=3y'(x+4y^2y')$ so that the derivative gets the factorized form $$ 0=3(y''y+2y'^2)(x+4y^2y'). $$ The first factor implies $y'y^2=C$ so that $y^3=3Cx+D$ and thus one obtains a linear family with $Y=y^3$.