Answer
True
Work Step by Step
Solve for $y$:
$\frac{dy}{dt}=2y(1-\frac{y}{5})$ (Separate the variables)
$\frac{dy}{dt}=\frac{2y(5-y)}{5}$
$\frac{5}{y(5-y)}dy=2dt$
$\frac{(5-y)+y}{y(5-y)}dy=2dt$
$(\frac{1}{y}+\frac{1}{5-y})dy=2dt$ (Integrate)
$\int (\frac{1}{y}+\frac{1}{5-y})dy=\int 2dt$
$\ln y-\ln (5-y)=2t+C$
Substitute the initial condition $y(0)=1$:
$\ln 1-\ln (5-1)=2\cdot 0+C$
$0-\ln 4=0+C$
$C=-\ln 4$
We have now:
$\ln y-\ln (5-y)=2t-\ln 4$
$\ln y-\ln (5-y)=\ln e^{2t}-\ln 4$
$\ln (\frac{y}{5-y})=\ln \frac{e^{2t}}{4}$
$\frac{y}{5-y}=\frac{e^{2t}}{4}$
$\frac{5-y}{y}=\frac{4}{e^{2t}}$
$\frac{5}{y}-1=4e^{-2t}$
Since as $t\to \infty$, $4e^{-2t}\to 0$, it must be $\frac{5}{y}-1\to 0$, or $y\to 5$.
Thus, $\lim\limits_{t \to \infty }y=5$ (TRUE).
