Question

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

Answer 1

We have $\frac{d y}{d x}=20\left(1600-y^{2}\right)$
$\Rightarrow \int \frac{d y}{(40)^{2}-y^{2}}=20 \int d x$
$\Rightarrow \frac{1}{80} \ln \frac{40+y}{40-y}=20 x +C'$
or $\ln \frac{40+y}{40-y}=1600 x+ C$
$\Rightarrow \frac{40+y}{40-y}=\frac{k e^{1600 x}}{1},$
where $k=e^{C}$ (let)
$\Rightarrow \frac{2 y}{80}=\frac{k e^{1600 x}-1}{k e^{1600 x}+1}$ (using componendo and dividendo)
$\therefore \displaystyle \lim_{x \rightarrow \infty} y=40 \displaystyle \lim_{x \rightarrow \infty}\left[\frac{k-e^{-1600 x}}{k-e^{-1600 x}}\right]=40$