so i know this question can be done both way

y' = 3y - y^2

<=>

y' = y(3-y)

alright so i dont think i can do it by bernoulli because i dont have two variables so i dont think thats gonna work but

I need to solve the initial value problem y(0) = 2

i get up to

dy

-- = 3dt

y(1-y/3)

i just don't know how to to setup and do the partial fraction