Answer to Question #121364 in Differential Equations for Gilbert Gentio

Question

y(2xy+1)dx-xdy=0

Answer 1

$y(2xy+1)dx-xdy=0\\ y(2xy+1)-x\frac{dy}{dx}=0\\ y'-\frac{1}{x}y = 2y^2$

it is Bernoulli equation

$v=y^{-1}\\ -v'-\frac{v}{x}=2\\ v'+\frac{v}{x} = 0\\ \frac{dv}{v} = -\frac{dx}{x}\\ \ln(v) = -\ln(x)+C\\ v(x) = \frac{C}{x}$

general solution is:

$v(x) = -x+\frac{c}{x}\\ y(x) = v^{-1}(x) = \frac{x}{C-x^2}$