Question

Given $y(0)=2000$ and $\frac{d y}{d x}=32000-20 y^2$, then find the value of $\underset{x \rightarrow \infty}{\operatorname{Lim}} y(x)$.

Answer 1

$ \text { We have } \frac{ dy }{ dx }=20\left(1600- y ^2\right) $
$\Rightarrow \int \frac{ dy }{(40)^2- y ^2}=20 \int dx$
$\Rightarrow \frac{1}{80} \ln \frac{40+ y }{40- y }=20 x + C ^{\prime} \text { or } \ln \frac{40+ y }{40- y }=1600 x + C$
$\Rightarrow \frac{40+ y }{40- y }=\frac{ ke ^{1600 x }}{1}, \text { where } k = e ^{ c } \text { (let) } $
$\Rightarrow \frac{2 y }{80}=\frac{ ke ^{1600 x }-1}{ ke ^{1600 x }+1} \text { (using componendo \& dividendo) } $
$\therefore \underset{x \rightarrow \infty}{\operatorname{Lim}} y =40 \underset{x \rightarrow \infty}{\operatorname{Lim}}\left[\frac{ k - e ^{-1600 x }}{ k - e ^{-1600 x }}\right]=40$