Solving a Linear Differential Equation

[misogynisticfeminist]

I've got a linear DE here,

$$(x + 4y^2) dy + 2y dx =0$$

I've tried to put it in the general form of a linear equation, and I would get,

$$\frac {dy}{dx} + \frac {2y}{x+4y^2} = 0$$

but I have problems isolating the x, so that I would get the $$P(x)$$ in the general form.

 

[AKG]

This doesn't make sense. If "DE" stands for "differential equation" then no, you don't have a DE here.

 

[misogynisticfeminist]

^ sorry, its supposed to be equals to 0, i forgot to add that...

 

[Hurkyl]

Think outside the box for a moment. There's a very easy way to turn this into an ODE.

 

[James R]

Your D.E. is not linear, either.

 

[HallsofIvy]

The way you have written it
$$\frac {dy}{dx} + \frac {2y}{x+4y^2} = 0$$
it is not linear.
However, you can write
$$(x + 4y^2) dy + 2y dx =0$$

as
$$\frac{dx}{dy}= -\frac{x+4y^2}{2y}= -\frac{1}{2y}x- 2y$$
which is a linear d.e. for x as a function of y.