## Calculus (3rd Edition)

$$y= \frac{5}{1+3e^{-3t}/2}.$$
By comparison with the logistic equation $\frac{d y}{d t}=k y\left(1-\frac{y}{A}\right)$ we have $k=3$ and $A=5$. Hence, the solution is given by $$y=\frac{A}{1-e^{-kt}/B}=\frac{5}{1-e^{-3t}/B}.$$ Since $y(0)=2$, then $2=\frac{5}{1-1/B}$ and hence $2=\frac{5B}{B-1}$. Then $2B-2=5B$ and $B=-\frac{2}{3}$ and $$y= \frac{5}{1+3e^{-3t}/2}.$$