the question is to solve the ordinary differential equation $x=py-p^2$ where p=$\left(\frac{dy}{dx}\right)$ I have tried the 3 methods of :


*

*solving for $x$,

*solving for $y$, and 

*solving for $p$, where $p$ is $\left(\frac{dy}{dx}\right)$.
In each of the above form, i could not get the solvable equation.
I am also not able to find the suitable substitution to modify it to the Clairaut's Form
I need a direction to proceed.
 A: Differentiate the given equation w.r.t $x$,
     $$\dfrac{dx}{dy} = p + y\dfrac{dp}{dy} - 2p\dfrac{dp}{dy}$$
     i.e. $$\dfrac{1}{p} = p + y\dfrac{dp}{dy} - 2p\dfrac{dp}{dy}$$
     From this we get a linear differential equation of the following form:
     $$\dfrac{dy}{dp} + \dfrac{p}{p^2 - 1}y = \dfrac{2p^2}{p^2 - 1},$$
     This is a linear ordinary differential equation of first order and first degree and can be solved easily.
A: $x=y\dfrac{dy}{dx}-\left(\dfrac{dy}{dx}\right)^2$
$\left(\dfrac{dy}{dx}\right)^2-y\dfrac{dy}{dx}+x=0$
Apply the method in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=2:
Let $F(x,y,t)=t^2-yt+x~,$
Then $\dfrac{dy}{dt}=-\dfrac{t\dfrac{\partial F}{\partial t}}{\dfrac{\partial F}{\partial x}+t\dfrac{\partial F}{\partial y}}=-\dfrac{t(2t-y)}{1+t(-t)}=\dfrac{2t^2}{t^2-1}-\dfrac{ty}{t^2-1}$
$\dfrac{dy}{dt}+\dfrac{ty}{t^2-1}=\dfrac{2t^2}{t^2-1}$
$y=t+\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{\sqrt{t^2-1}}$
$\therefore\dfrac{dx}{dt}=-\dfrac{\dfrac{\partial F}{\partial t}}{\dfrac{\partial F}{\partial x}+t\dfrac{\partial F}{\partial y}}=-\dfrac{2t-y}{1+t(-t)}=\dfrac{2t}{t^2-1}-\dfrac{y}{t^2-1}=\dfrac{t}{t^2-1}-\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{(t^2-1)^\frac{3}{2}}$
$x=\int\biggl(\dfrac{t}{t^2-1}-\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{(t^2-1)^\frac{3}{2}}\biggr)dt=\dfrac{t(\ln(t+\sqrt{t^2-1})+C_1)}{\sqrt{t^2-1}}+C_2$
Hence $\begin{cases}x=\dfrac{t(\ln(t+\sqrt{t^2-1})+C_1)}{\sqrt{t^2-1}}+C_2\\y=t+\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{\sqrt{t^2-1}}\end{cases}$
A: First differentiate w.r.t $x$ ,we get
$$\frac{dx}{dy}=p+y\frac{dp}{dy} -2p \frac{dp}{dy}$$
Now solve for $\frac{dy}{dp}$ using linear differential equation 
A: The equation $F(x,y,p)=py-p^2-x$ can be solved for $p$ outside the singular set $0=F_p=y-2p$, that is, $(x,y)=(p^2,2p)$. This results in the differential equations
$$
y^2-4x=(2p-y)^2\implies y'=p=\frac12\left(y\pm\sqrt{y^2-4x}\right)
$$
which can be used to obtain numerical solutions. No integration tricks are immediately obvious.

One other approach is to use $p$ as parameter of the solutions wherever $p=y'$ is not constant.
For the lines $y'=p=c=const.$ the given equation results in $x=cy-c^2$. By differentiation this gives the condition $1=cy'=c^2$ so that only $c=\pm 1$ give solutions.
Outside these lines one can take $p$ as parameter, denote $\dot x=\frac{dx}{dp}$, $\dot y=\frac{dy}{dp}$. By the chain rule, $\dot y=p\dot x$. Taking the derivative by $p$ of the original equation and multiplying with $p$ gives
$$
\dot y=p\dot x=p(y+p\dot y-2p)\implies \dot y=\frac{py-2p^2}{1-p^2}
$$

For $|p|<1$ parametrize $p=\sin u$ so that
$$
\frac{d}{du}(\cos u\, y(\sin u))=\cos^2u\,\dot y(\sin u)-\sin u\, y(\sin u)=-2\sin^2u=\cos(2u)-1
$$
which integrates to
$$
\cos u y(\sin u)=C+\sin u\cos u-u.
$$
This gives, inserting this last equation in the original ODE, a parametrization
$$
\pmatrix{x\\y}=\pmatrix{(C-u)\tan u\\\sin u+\frac{C-u}{\cos u}}, ~~~ u\in(-\frac\pi2,\frac\pi2)
$$

For $p>1$ (and similarly for $p<-1$) parametrize $p=\cosh u$ so that $1-p^2=-\sinh^2u$ and
$$
\frac{d}{du}(\sinh u\,y(\cosh u))=\sinh^2u\,\dot y(\cosh u) + \cosh u y(\cosh u)
=2\cosh^2u=\cosh(2u)+1
$$
which integrates to
$$
\sinh u\,y(\cosh u)=C+u+\sinh u\,\cosh u.
$$
This gives, inserting this last equation in the original ODE, a parametrization
$$
\pmatrix{x\\y}=\pmatrix{(C+u)\coth u\\\cosh u+\frac{C+u}{\sinh u}}, ~~~ u\in(0,\infty)
$$

Comparing the parametrizations with numerical solutions confirms their correctness

Plotted in dark and light gray are the first and second parametrizations, red and blue are numerical solutions for both signs of the square root. Note that while the parametrizations pass resp. "are reflected at" the singular set (in yellow), the numerical solutions end there, there is no differentiable extension available. 
