How to solve tougher Clairaut's equation Since I have been finding all the answers here, so I am asking you all to help me with this one also:- 
Basically, my teacher / professor ask me to mug up all these, please can you help me find a general or any type of hints with which I can be able to solve these equations ::


*

*For $$ y=2xp + y^2 p^3 $$ $$ y^2 =v , x=u $$ 

*For $$ y+px =x^4 p^2 $$ --> $$ (1/x) = u, y=v $$ 

*For $$(p^2  + 1 ) (x-y) ^2 = (x+yp) ^2 $$ , $$ x=r cosθ , y= r sinθ $$ 

*For $$ x^2 p^2 + 2xyp + y^2 (1+p) =0 $$ $$ xy =u, y=v $$


And the list goes on. 
$p$ here means $dy/dx$; $y$ being a function of $x$ alone
What to do. Please guide here. 
 A: Solving ODEs of these kind requires


*

*First, to find some changes of variables $(x,y(x))\to (u,v(u))$ so that the ODE be transformed to a Clairaut's ODE.

*Second, perform the calculus of changes of variables leading to the Clairaut's ODE on the form :
$$v=u\frac{dv}{du}+f\left(\frac{dv}{du}\right)$$

*Third, solve the Clairaut's ODE, which is easy :
$$v=c\:u+f(c)$$

*Fourth, coming back to the original variables with the inverse change $(u,v(u))\to (x,y(x))$.


In fact, the exercice doesn't concern the first step since the changes of variables (of course different for each ODE) are given in the wording. The exercice is a training for the second, third and fourth steps only. This is an exercice adapted to the level of the students.
The first step is far the most difficult. In general, it is so difficult that it isn't adapted to the level of the students. That is why the convenient changes of variables are given (except in some elementary textbook cases).
As a matter of fact, there is no general method to find the changes of variables convenient to a given ODE. Intuition, Experience, Trial and Error, etc. are usual approaches. Don't worry, if you are faced to such an exercice without given changes of variables, this means that the changes of variables are very simple, almost obvious, and adapted to your level of knowledge so that you are able to find them.
A: You can also just consider all functions as functions in $p$ with the additional relation $\dot y=p\dot x$ and then try to isolate one of $x,\dot x$ or $y,\dot y$. In your examples this works as follows


*

*In $y=2xp+y^2p^3$ you can  take the $p$ derivative then eliminate $x$ to get
$$ \dot y=2\dot xp+2x+2y\dot yp^3+3y^2p^2
\\ \dot y=2\dot y+\frac1p(y-y^2p^3)+2y\dot yp^3+3y^2p^2
\\     0 =\dot y+\frac yp +2y\dot yp^3+2y^2p^2=\left(\dot y+\frac yp\right)\left(1+2yp^3\right)$$
which has solutions $py=C$ for the first factor and $y=-\frac1{2p^3}$ for the second factor. This both can be inserted into the first equation to get x as function of $p$, after that one can attempt to eliminate $p$. Or you could recognize $C=py=y'(x)y(x)=\frac12\frac{d(y^2)}{dx}$ for $y^2=ax+b$ in the first factor and then try similarly to substitute $v=y^2$ in the second factor.

*$y+px=x^4p^2$. Take the $p$ derivative and eliminate $y$
$$ \dot y+p\dot x+x=4x^3\dot xp^2+2x^4p 
\\2p\dot x +x=2x^3p(2p\dot x+x)
\\0=(2p\dot x+x)(2x^3p-1)$$
The first factor gives $px^2=C$ and the second $px^3=\frac12$ which both allows for a direct elimination of $p$ from the original equation.
etc. That such a simplifying factorization occurs is due to the construction of the problems as modified Clairaut equations, almost any change in the coefficients will destroy that pattern.
