Question

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

$\frac{d y}{d x}+y=\frac{1}{1+e^{2 x}}$
So integrating factor is $e^{\int 1 . d x}=e^x$
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 \displaystyle\lim _{x \rightarrow \infty}\left(y \cdot e^x\right)= \displaystyle\lim _{x \rightarrow \infty}\left(\tan ^{-1}\left(e^x\right)+\frac{\pi}{4}\right)=\frac{3 \pi}{4}$