Question

Given $y(0)=2000$ and $\frac{dy}{dx}=32000-20y^2$, then find the value of $\underset{x\to \infty }{\lim }y(x)$.

Answer 1

Answer: 40

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