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 $k$. Then what is the value of $5k?$

Answer 1

$\frac{dy}{dx}=25y^2-4$
So, $\frac{dy}{25y^2-4}=dx$
Integrating, $\frac{1}{25}\times \frac{1}{2\times \frac{2}{5}}ln\left|\frac{y-\frac{2}{5}}{y+\frac{2}{5}}\right|=x+c$

$\Rightarrow ln\left|\frac{5y-2}{5y+2}\right|=20(x+c)$

Now, $c=0$ as $f\left(0\right)=0$

Hence, $\left|\frac{5y-2}{5y+2}\right|=e^{\left(20x\right)}$

$\therefore \underset{x\to -\infty }{lim}\left|\frac{5f\left(x\right)-2}{5f\left(x\right)+2}\right|=\underset{x\to -\infty }{lim}{e}^{\left(20x\right)}$

Now, $RHS=0\Rightarrow \underset{x\to -\infty }{lim}\left(5f\right(x)-2)=0$

$\Rightarrow \underset{x\to -\infty }{lim}f\left(x\right)=\frac{2}{5}=0.40$.