Question

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

• A. 1
• B. -1
• C. 0
• D. $\frac{2}{2-e}$

Answer 1

The correct option is A 1

$\frac{dy}{dx}=y-y^2$ $y\left(0\right)=2$

$\frac{dy}{y-y^2}=dx$

$\frac{-dy}{y^2-y}=dx$

$\frac{-dy}{y^2{\left(\frac{1}{2}\right)}^2-2\times \frac{1}{2}y-\frac{1}{4}}=dx$

$\frac{dy}{{(y-\frac{1}{2})}^2-{\left(\frac{1}{2}\right)}^2}=-dx$

$\frac{1}{2\times \frac{1}{2}}\ln \left|\frac{y-\frac{1}{2}-\frac{1}{2}}{y-\frac{1}{2}+\frac{1}{2}}\right|=-x+c$

$\ln \left(\frac{y-1}{y}\right)=x+c$

Given $y\left(0\right)=2$

$\ln \left(\frac{2-1}{2}\right)=C$

$C=\ln \left(\frac{1}{2}\right)$

$\ln \left(\frac{y-1}{y}\right)=-x+\ln \left(\frac{1}{2}\right)$

$\ln \left(\frac{y}{y-1}\right)=x+\ln \left(2\right)$

$\frac{y}{y-1}=e^{x+\ln \left(2\right)}$

$\frac{y-1}{y}=\frac{1}{e^{x+\ln \left(2\right)}}$

$1-\frac{1}{y}=e^{-(x+\ln (2\left)\right)}$

$\frac{1}{y}=1-e^{-(x+\ln (2\left)\right]}$

$y=\frac{1}{1-e^{-[x+\ln (2\left)\right]}}$

$\underset{x\to \infty }{lim}\frac{1}{1-e^{-\infty }}=\frac{1}{1-0}=1$