Let the solution curve y = y (x) of the differential equation (1 +...

Given,

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

Now on rearranging we get,

$⇒ \dfrac{dy}{dx} + y = \dfrac{1}{1 + {e}^{2x}}$

We can see it is a linear differential equation,

So integrating factor is ${e}^{∫ 1 \cdot dx} = {e}^{x}$

So solution will be

$y \cdot IF = ∫ \dfrac{1}{1 + {e}^{2x}} \times IFdx$

$⇒ y{e}^{x} = ∫ \dfrac{1}{1 + {e}^{2x}} \times {e}^{x}dx$

$⇒ y \cdot {e}^{x} = \tan^{- 1}\left({e}^{x}\right) + c$

Now as curve is passing through \left(0, \dfrac{\pi}{2}\right)$so </p><p>$⇒ c = \dfrac{\pi}{4}$</p><p>Now calculating the limit$\lim_{x \to \infty}\left(y \cdot {e}^{x}\right) we get,

$⇒ \lim_{x \to \infty}\left(y \cdot {e}^{x}\right)$ = \lim_{x \to \infty}\left(\tan^{- 1}\left({e}^{x}\right)$+$\dfrac{\pi}{4}\right) =\dfrac{\pi}{2} + \dfrac{\pi}{4} = \dfrac{3\pi}{4}