## Calculus (3rd Edition)

$$y= \frac{3}{1-7e^{-6t}/10}.$$
Rewriting the logistic equation as $\frac{d y}{d t}=6 y\left(1-\frac{y}{3}\right)$, we have $k=6$ and $A=3$ and hence the solution is given by $$y=\frac{A}{1-e^{-kt}/B}=\frac{3}{1-e^{-6t}/B}.$$ Since $y(0)=10$, then $10=\frac{3}{1-1/B}$ and hence $10=\frac{3B}{B-1}$. Then $10B-10=3B$ and $B=\frac{10}{7}$ and $$y= \frac{3}{1-7e^{-6t}/10}.$$