Question

Let $f:R\to R$ be a differentiable function with $f\left(0\right)=0.$ If $y=f\left(x\right)$ satisfies the differential equation
$\frac{dy}{dx}=(2+5y)(5y-2)$
then the value of $\underset{x\to -\infty }{lim}f\left(x\right)$ is

Answer 1

$\frac{dy}{dx}=25y^2-4\Rightarrow \frac{1}{25}.\frac{dy}{y^2-(\frac{2}{5}{)}^2}=dx$
Intregrating both side we get,
$\frac{1}{25}\frac{1}{2\times \frac{2}{5}}\ln |\frac{y-\frac{2}{5}}{y+\frac{2}{5}}|=x+c$
Now, $c=0$ as $f\left(0\right)=0.$
Hence, $\frac{5y-2}{5y+2}=e^{20x}\Rightarrow y=f\left(x\right)=\frac{2}{5}|\frac{e^{20x}-1}{e^{20x}+1}|$
$\underset{x\to -\infty }{lim}f\left(x\right)=\underset{x\to -\infty }{lim}\frac{2}{5}|\frac{e^{20x}-1}{e^{20x}+1}|=\frac{2}{5}=0.4$