Discussion Forum

1)

Let $f:R\rightarrow R$ be a differentiable function with f(0)=0,  y=f(x) satisfies the differential equation

 $\frac{\text{d}y}{\text{d}x}= (2+5y)(5y-2)$  , then the value of  $\lim_{x \rightarrow \infty} f(x)$  is .............

Answer:

Option B

Explanation:

We have , $\frac{\text{d}y}{\text{d}x}=(2+5y)(5y-2)$

$\Rightarrow$     $\frac{dy}{25y^{2}-4}=dx$

$\Rightarrow$     $\frac{1}{25}(\frac{dy}{y^{2}-\frac{4}{25}})=dx$

  On integrating both sides, we get ,$\frac{1}{25}\int_{}^{} \frac{dy}{y^{2}-(\frac{2}{5})^{2}}=\int_{}^{} dx$

$\Rightarrow$    $\frac{1}{25}\times\frac{1}{2\times 2/5}\log_{}{\mid\frac{y-2/5}{y+2/5}}\mid =x+C$

$\Rightarrow$  $log_{}{\mid\frac{5y-2}{5y+2}}\mid =20(x+C)$

$\Rightarrow$     ${\mid\frac{5y-2}{5y+2}}\mid =Ae^{20x}$    [ $\therefore$       $e^{20C}=A$ ]

when x=o    $\Rightarrow$  y =0 then A =1

$\therefore$           $\mid \frac{5y-2}{5y+2}\mid =e^{20x}$

 $\lim_{x \rightarrow -\infty} \mid \frac{5f(x)-2}{ 5f(x)+2}\mid =\lim_{x \rightarrow -\infty}e^{20x}$

$\Rightarrow$   $\Rightarrow \lim_{n \rightarrow -\infty} \mid \frac{5f(x)-2}{ 5f(x)+2}\mid =0$

$\Rightarrow  \lim_{n \rightarrow -\infty} 5 f(x) -2=0$

$\Rightarrow  \lim_{n \rightarrow -\infty} f(x) =\frac{2}{5}=0.4$