Solving an implicit first-order differential equation

Question

We have a problem stating:

Solve $ y(6y^2-x-1)dx + 2xdy = 0 $

Since we can't simply separate the variables. Our theory state we can use theses formulas to find a factor that's only dependent on a single variable:

But trying both cases gives us Integrating factors that is dependent on two variables:

Case 1: $ f(x) = \frac {-(x-3(6y^2-1))}{2x} $

Case 2: $ g(y) = \frac {x-3(6y^2-1)}{(x-6y^2+1)y} $

We're not sure what to do next or if we even used the correct method for solving this.

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EDIT: Here's the full solution using the accepted answer from @user577215664.

Answer 1

Hint: $$y(6y^2-x-1)dx + 2xdy = 0$$ $$-2xy'=y(6y^2-x-1) $$ $$2xy'=y(x+1)-6y^3 $$ This is Bernoulli's differential equation. Maybe you have seen how to solve this specific DE in your course ?

Answer 2

Here is a solution which relies on trial and error, looking for substitutions which appear likely to simplify the equation. First, to get rid of the $y^2$, substitute $$u=y^2\ ,\quad \frac{du}{dx}=2y\,\frac{dy}{dx}$$ and do the algebra carefully to get $$(6u-x-1)+\frac{x}{u}\,\frac{du}{dx}=0\ .$$ Now substitute $v=u/x$ and simplify to give $$\frac{dv}{dx}=v-6v^2\ .$$ This is separable so I think I can leave it to you from here. My final answer was $$y^2(1+6Ae^x)=Axe^x\ .$$