Solve this equation using Bernoulli's equation How do you solve this equation: 
$$\frac{dy}{dx} + \frac{1}{xy} = y^3$$
using Bernoulli's equation.
I've tried everything but I think there may be an error with the actual equation.
 A: Your equation has the form of $y'+P(x)y^{-1}=Q(x)y^n$, to get the equation lulu stated is impossible from here. (note the power of -1)
A: Let $u=y^2$ ,
Then $\dfrac{du}{dx}=2y\dfrac{dy}{dx}$
$\therefore\dfrac{1}{2y}\dfrac{du}{dx}+\dfrac{1}{xy}=y^3$
$\dfrac{du}{dx}=2y^4-\dfrac{2}{x}$
$\dfrac{du}{dx}=2u^2-\dfrac{2}{x}$
Which reduces to a Riccati ODE.
Let $u=-\dfrac{1}{2v}\dfrac{dv}{dx}$ ,
Then $\dfrac{du}{dx}=\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{1}{2v}\dfrac{d^2v}{dx^2}$
$\therefore\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{1}{2v}\dfrac{d^2v}{dx^2}=2\left(-\dfrac{1}{2v}\dfrac{dv}{dx}\right)^2-\dfrac{2}{x}$
$\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{1}{2v}\dfrac{d^2v}{dx^2}=\dfrac{1}{2v^2}\left(\dfrac{dv}{dx}\right)^2-\dfrac{2}{x}$
$\dfrac{1}{2v}\dfrac{d^2v}{dx^2}=\dfrac{2}{x}$
$x\dfrac{d^2v}{dx^2}-4v=0$
$v=C_1\sqrt{x}I_1(4\sqrt{x})+C_2\sqrt{x}K_1(4\sqrt{x})$ (e.g. according to https://www.wolframalpha.com/input/?i=x+v%22-4v%3D0)
$\therefore u=-\dfrac{\dfrac{d}{dx}(C_1\sqrt{x}I_1(4\sqrt{x})+C_2\sqrt{x}K_1(4\sqrt{x}))}{2C_1\sqrt{x}I_1(4\sqrt{x})+2C_2\sqrt{x}K_1(4\sqrt{x})}$
$y^2=-\dfrac{2C_1I_0(4\sqrt{x})-2C_2K_0(4\sqrt{x})}{2C_1\sqrt{x}I_1(4\sqrt{x})+2C_2\sqrt{x}K_1(4\sqrt{x})}$ (according to https://www.wolframalpha.com/input/?i=d%2Fdx(x%5E(1%2F2)besseli(1,4x%5E(1%2F2)) and https://www.wolframalpha.com/input/?i=d%2Fdx(x%5E(1%2F2)besselk(1,4x%5E(1%2F2)))
$y^2=-\dfrac{I_0(4\sqrt{x})-CK_0(4\sqrt{x})}{\sqrt{x}I_1(4\sqrt{x})+C\sqrt{x}K_1(4\sqrt{x})}$
