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True/False Question :

Let $y\left ( t \right )$ be a real valued function defined on the real line such that ${y}'=y \left ( 1-y \right )$, with $y\left ( 0\right ) \in \left [ 0,1 \right ]$. Then $\lim_{t\rightarrow \infty }y\left ( t \right )=1$ .

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$\text{$\frac{dy}{dt} = y(1-y)$}$

$\text{$\int \frac{dy}{y(1-y)} =\int dt$}$

$\text{$\int \left ( \frac{1}{y} + \frac{1}{1-y} \right )dy = \int dt$}$

$\text{$\log |y| \; –\; \log |(1-y)| +c = t$ (c is arbitrary constant)}$

$\text{$\log y \; –\; \log (1-y)  = t – c$}$

$\text{$\log \frac{y} {(1-y)}  = t – c$}$

$\text{$ \frac{y} {1-y}  =e^{ t – c}$}$

$\text{$ y= \frac{e^{t-c}}{1+e^{t-c}}= \frac{1}{1+e^{c-t}} = \frac{1}{1+\frac{e^{c}}{e^t}}$}$  
As $t \rightarrow \infty, e^t \rightarrow \infty,$ So, $\lim_{t \rightarrow \infty } y(t) = 1$

$\text{But there are 2 }$ trivial $\text{solutions also for the given differential equation i.e. $y(t)=0$ and $y(t)=1$  }$

$\text{which satisfies the given differential equation $y’=y(1-y)$}$

$\text{and so, $\lim_{t \rightarrow \infty } y(t) =  \lim_{t \rightarrow \infty } 0 = 0$}$

$\text{So, given statement is False.}$

Answer:

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