Question

Xp^2+2yp+x=0

Answer 1

$xp^2+2yp+x = 0\\ 2y = -xp - \dfrac{x}{p}\\ 2\dfrac{dy}{dx} = -p -\dfrac{1}{p} -x\dfrac{dp}{dx}+\dfrac{x}{p^2}\dfrac{dp}{dx}\\ \dfrac{3p^2+1}{p} = \dfrac{x(1-p^2)}{p^2}\dfrac{dp}{dx}\\ \dfrac{(1-p^2)}{p(3p^2+1)}dp = \dfrac{dx}{x}\\ \int \dfrac{(1-p^2)}{p(3p^2+1)}dp =\int \dfrac{dx}{x}\\ lnp -\dfrac{2}{3}ln(3p^2+1) = lnx\\ \dfrac{p}{\sqrt[3]{9p^4+6p^2+1}} = x\\ 9x^3p^4-p^3+6x^3p^2+x^3=0$

is an implicit solution of the question because it is impossible to express p straightforward in terms of y.