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