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

$\begin{array}{l}Given\frac{dy}{dx}=y-y^2\\ \Rightarrow \int \frac{1}{y(1-y)}dy=\int dx\\ \Rightarrow \int \frac{(1-y)+y}{y(1-y)}dy=\int dx\\ \Rightarrow \int (\frac{1}{y}+\frac{1}{1-y})dy=\int dx\\ \Rightarrow \int \frac{1}{y}dy-\int \frac{1}{y-1}dy=\int dx\\ \Rightarrow \log y-\log (y-1)=x+c\\ \Rightarrow \log \left(\frac{y}{y-1}\right)=x+c\\ Putx=0\Rightarrow \log \left(\frac{y\left(0\right)}{y\left(0\right)-1}\right)=c\\ \Rightarrow c=\log 2\end{array}$

$\begin{array}{l}\therefore \log \left(\frac{y}{y-1}\right)=x+\log 2\\ \Rightarrow \log \frac{y}{2(y-1)}=x\\ \Rightarrow \frac{y}{2(y-1)}=e^x\\ \Rightarrow y=2y{e}^x-2e^x\\ \Rightarrow 2y{e}^x-y=2e^x\\ \Rightarrow y(2e^x-1)=2e^x\\ \Rightarrow y=\frac{2e^x}{2e^x-1}\\ \Rightarrow y=\frac{2}{2-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e^x$}\right.}\\ \underset{x\to \infty }{Lt}y\left(x\right)=\underset{x\to \infty }{Lt}\left(\frac{2}{2-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$e^x$}\right.}\right)\\ =\left(\frac{2}{2-0}\right)\Rightarrow 1\end{array}$