What is the Bernoulli Differential Equation Form of 3y^2y' + y^3 = e^-x?

[Sean77771]

The Problem:

3y²y' + y³ = e^-x

I think maybe I'm supposed to use something to do with the Bernoulli stuff, but I'm not sure. I've tried to figure it out for a while now and I'm stuck.

Thanks for any help you can give.

 

[Defennder]

It's in the form of a Bernoulli DE if you divide everything by 3y^2.