Question

The value of $\underset{x\to \infty }{lim}y\left(x\right)$ obtained from the differential equation $\frac{dy}{dx}=y-y^2,$ where $y\left(0\right)=2$ is

Answer 1

$\frac{dy}{dx}=y-y^2$
$\Rightarrow \frac{dy}{y(y-1)}=-dx$
Integrating both sides, we have
$\int \frac{dy}{y(y-1)}=-\int dx\Rightarrow \int (\frac{1}{y-1}-\frac{1}{y})dy=-\int dx\Rightarrow \ln \left|\frac{y-1}{y}\right|=-x+c$
Given, $y\left(0\right)=2\Rightarrow c=-\ln 2$
$\therefore \ln \left|\frac{y-1}{y}\right|=-x-\ln 2$
$\Rightarrow \left|\frac{2(y-1)}{y}\right|=e^{-x}$
When $x\to \infty ,$ we get
$\left|\frac{2(y-1)}{y}\right|=0$
$\Rightarrow y=1$