the question is to solve the ordinary differential equation $x=py-p^2$ where p=$\left(\frac{dy}{dx}\right)$

Question

I have tried the 3 methods of :

1. solving for $x$,

2. solving for $y$, and

3. 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.

Answer 1

$x=y\dfrac{dy}{dx}-\left(\dfrac{dy}{dx}\right)^2$

$\left(\dfrac{dy}{dx}\right)^2-y\dfrac{dy}{dx}+x=0$

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 $\begin{cases}x=\dfrac{t(\ln(t+\sqrt{t^2-1})+C_1)}{\sqrt{t^2-1}}+C_2\\y=t+\dfrac{\ln(t+\sqrt{t^2-1})+C_1}{\sqrt{t^2-1}}\end{cases}$