Log Issues in Differential Equation

Question

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. $$\ln y-\ln(3-y)=3(t^2/2+c)$$ $$\ln[y/(3-y)]=3(t^2/2+c)$$ $$e^{\ln[y/(3-y)]}=e^{3(t^2/2+c)}$$ $$y/(3-y)=e^{3(t^2/2+c)}$$ $$(3y^{-1}-1)^{-1}=e^{3(t^2/2+c)}$$ $$3y^{-1} -1=e^{-3(t^2/2+c)}$$ $$3y^{-1}=e^{-3(t^2/2+c)}+1$$ $$y^{-1}=1/3(e^{-3(t^2/2+c)}+1)$$ $$y=3(e^{-3(t^2/2+c)}+1)^{-1}$$

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

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Where am I messing up?

Answer 1

You got to \begin{eqnarray*} y=3(e^{-3(t^2/2+c)}+1)^{-1}. \end{eqnarray*} Now multiply top & 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*}

Answer 2

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: $$y \left( t \right) =3\, \left( 1+3\,{{\rm e}^{-3/2\,{t}^{2}}}{\it \_C1 } \right) ^{-1} $$