Question

Let the solution curve $y=y$ (x) of the differential equation $\left(1+e^{2x}\right)\left(\frac{dy}{dx}+y\right)=1$ pass through the point $\left(0,\frac{\pi }{2}\right)$. Then, $\underset{x\to \infty }{\lim }$ $e^{x}y(x)$ is equal to :

• A. $\frac{\pi }{4}$
• B. $\frac{3\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\frac{3\pi }{2}$

Answer 1

Answer: (B)

$\frac{dy}{dx}+y=\frac{1}{1+e^{2x}}$
So integrating factor is $e^{\int 1.dx}=e^x$
So solution is $y\cdot e^x={\tan }^{-1}\left(e^x\right)+c$
Now as curve is passing through
$\left(0,\frac{\pi }{2}\right)\text{ so }$
$\Rightarrow c=\frac{\pi }{4}$
$\Rightarrow \underset{x\to \infty }{\lim }\left(y\cdot e^x\right)=\underset{x\to \infty }{\lim }\left({\tan }^{-1}\left(e^x\right)+\frac{\pi }{4}\right)=\frac{3\pi }{4}$