Solve the second degree differential equation $(1- y^2+ \frac{y^4}{x^2})p^2-\frac{2yp}{x}+\frac{y^2}{x^2}=0$

Question

Solve the second degree differential equation.

$(1- y^2+ \frac{y^4}{x^2})p^2-\frac{2yp}{x}+\frac{y^2}{x^2}=0$ where $p=dy/dx$

I tried to solve as follows but got stuck. Please help. Other “good” ways to solve this equation are also welcome.

Answer 1

Far from a solution, a small advance that may inspire you:

Let us consider the half quadrant $x>y>0$ for which you have:

$$\frac{dy}{dx}=\dfrac{y}{x-y\sqrt{x^2-y^2}}\tag{*}$$

Convert this differential equation into the equivalent differential system:

$$\begin{cases}\frac{dy}{dt}&=&-y& \ \ Eq. 1\\ \frac{dx}{dt}&=&-x+y\sqrt{x^2-y^2}& \ \ Eq. 2 \end{cases}$$

Eq. 1 gives $y=ae^{-t}$. Plugging this expression in Eq. 2, one gets:

$$\frac{dx(t)}{dt}=-x+ae^{-t}\sqrt{x^2-a^2e^{-2t}}$$

which has $x(t)=ae^{-t}$ as a particular solution.

Having $y=x$ as a solution is confirmed by the vector field associated with initial equation (*).