Use the separation of variables to find the general solution to $y' = 3y - y^2$.
Given $$\frac{dy}{dt} = 3y - y^2,$$ I can rearrange this to $$\frac{1}{3y-y^2}dy = dt$$
and using partial fraction decomposition
where $$\frac{1}{3y-y^2} = \frac{A}{y} + \frac{B}{3-y},$$
I get that $$A = B = \frac{1}{3}$$.
Then I have $$(\frac{1}{3y} - \frac{1}{3(3-y)})dy = dt$$
Integrating both sides, $$\frac{1}{3} \ln(|y|) + \frac{1}{3} \ln(|3-y|) = t+c_1,$$
which can also be written as $$\ln(|y(3-y)|) = 3t + c_2$$ $$\rightarrow |y(3-y)| = Ce^{3t}$$
I guessed I could solve for $y$ in this quadratic equation, but that doesn't give me the right solutions (which I have access to). What am I doing wrong?
You made some sign mistakes. Here I suppose $y \ne 0, y \ne 3$ $$\frac {dy}{y(3-y)}=dx$$ $$\frac {dy}{y}+\frac {dy}{3-y}=3dx$$ $$\frac {dy}{y}-\frac {dy}{(y-3)}=3dx$$ $$\ln| \frac {y}{(y-3)}|=3x+K$$ $$| \frac {y}{(y-3)}|=Ce^{3x}$$ $$ y(1-Ce^{3x})=-3Ce^{3x}$$ Rearrange terms you should get the right answer $$ y=\frac {-3Ce^{3x}}{(1-Ce^{3x})}$$ $$ y=\frac {3Ke^{3x}}{(1+Ke^{3x})}=\frac {3}{(1+Ce^{-3x})}$$