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 $\underset{x\to \mathrm{\infty }}{lim} f\left(x\right)$ is

We have,

$\begin{array}{l}\frac{dy}{dx}=(2+5y)(5y-2)\\ \Rightarrow \frac{1}{(5y+2)(5y-2)}dy=dx\\ \Rightarrow \frac{1}{4}\left(\frac{1}{5y-2}-\frac{1}{5y+2}\right)dy=dx\end{array}$

${\int }_4^1 \int \left(\frac{1}{5y-2}-\frac{1}{5y+2}\right)dy=\int dx$

$\begin{array}{l}\Rightarrow \frac{1}{20}\log \left|\frac{5y-2}{5y+2}\right|=x+C\\ \Rightarrow \left|\frac{5y-2}{5y+2}\right|=e^{2\alpha (x+C)}\end{array}$

It is given that $f\left(0\right)=0$ i.e. $y=0$ at $x=0$

Putting this value in (i), we obtain

$\begin{array}{l}\frac{5y-2}{5y+2}=e^{20x}\\ \Rightarrow \frac{5y-2}{5y+2}=\pm e^{20x}\\ \Rightarrow \frac{10y}{-4}=\frac{1\pm e^{20x}}{\pm e^{20x}-1}\\ \Rightarrow \frac{5y}{2}=\frac{1\pm e^{20x}}{-1\pm e^{20x}}\\ \Rightarrow y=\frac{2}{5}\left(\frac{1\pm e^{20x}}{-1\pm e^{20x}}\right)\end{array}$

$\begin{array}{l}\Rightarrow \underset{x\to x}{lim} y=\frac{2}{5}\underset{x\to x}{lim} \left(\frac{1\pm e^{20x}}{-1\pm e^{20x}}\right)\\ \Rightarrow \underset{x\to x}{lim} f\left(x\right)=\frac{2}{5}\left(\frac{0\pm 1}{0\pm 1}\right)=\frac{2}{5}\end{array}$