Log Issues in Differential Equation So, I'm trying to solve the following DE:
$$y'=yt(3-y)$$

And I've gotten to:
$$1/3(\ln y-\ln(3-y))=t^2/2+c$$
But, I'm having trouble figuring out why my solution isn't matching up.
$$\ln y-\ln(3-y)=3(t^2/2+c)$$
$$\ln[y/(3-y)]=3(t^2/2+c)$$
$$e^{\ln[y/(3-y)]}=e^{3(t^2/2+c)}$$
$$y/(3-y)=e^{3(t^2/2+c)}$$
$$(3y^{-1}-1)^{-1}=e^{3(t^2/2+c)}$$
$$3y^{-1} -1=e^{-3(t^2/2+c)}$$
$$3y^{-1}=e^{-3(t^2/2+c)}+1$$
$$y^{-1}=1/3(e^{-3(t^2/2+c)}+1)$$
$$y=3(e^{-3(t^2/2+c)}+1)^{-1}$$



And the solution should be:
$$y=3(e^{3(t^2/2+c)})(e^{3(t^2+c)}+1)^{-1}$$



Where am I messing up?
You got to 
\begin{eqnarray*}
y=3(e^{-3(t^2/2+c)}+1)^{-1}.
\end{eqnarray*}
Now multiply top &amp; bottom of the RHS by $e^{3(t^2/2+c)}$ and we have
\begin{eqnarray*}
y=3e^{3(t^2/2+c)}(1+e^{3(t^2/2+c)})^{-1} \\
y=\frac{3e^{3(t^2/2+c)}}{(1+e^{3(t^2/\color{red}{2}+c)})}. \\
\end{eqnarray*}
I think the mistake is from

$$\frac{y}{y-3}=e^{\frac{3}{2}t^2+3C}$$ to the next line. Multiplying by $$y-3$$ you will get

$$y=(y-3)A$$ so
$$y=\frac{3A}{A-1}$$ where $$A=e^{\frac{3}{2}t^2+3C}$$
For your Control: here is the Maple solution:
$$y \left( t \right) =3\, \left( 1+3\,{{\rm e}^{-3/2\,{t}^{2}}}{\it \_C1
} \right) ^{-1}
$$