Question

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

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

Answer 1

The correct option is B $\frac{3\pi }{4}$

$\text{D.E.}(1+e^{2x})(\frac{dy}{dx}+y)=1$

$\Rightarrow \frac{dy}{dx}+y=\frac{1}{1+e^{2x}}$

$\text{I.F.}=e^{\int 1.dx}=e^x$

$\therefore$ Solution

$e^{x}y\left(x\right)=\int \frac{e^x}{1+e^{2x}}dx$

$\Rightarrow e^{x}y\left(x\right)={\tan }^{-1}\left(e^x\right)+C$

$\because$ it passes through $(0,\frac{\pi }{2}),C=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$

$\therefore \underset{x\to \infty }{lim}{e}^{x}y\left(x\right)=\underset{x\to \infty }{lim}{\tan }^{-1}\left(e^x\right)+\frac{\pi }{4}$

$=\frac{3\pi }{4}$