If y = y (x) is the solution of the differential equation (1 + e ^...

Given,

$\left(1 + {e}^{2x}\right)\dfrac{dy}{dx} + 2\left(1 + {y}^{2}\right){e}^{x} = 0$

$⇒ \dfrac{dy}{1 + {y}^{2}} + \dfrac{2{e}^{x}}{1 + {e}^{2x}}dx = 0$ .........(i)

Now integrating both side we get,

$⇒ ∫ \dfrac{dy}{1 + {y}^{2}} + ∫ \dfrac{2{e}^{x}}{1 + {e}^{2x}}dx = ∫ 0$

$⇒ \tan^{- 1}y + 2\tan^{- 1}{e}^{x} = c$

$∵ y\left(0\right) = 0$

so, $C = \dfrac{\pi}{2} ⇒ \tan^{- 1}y + 2\tan^{- 1}{e}^{x} = \dfrac{\pi}{2}$ ....(ii)

Now from equation (i), we get ${\left(\dfrac{dy}{dx}\right)}_{x = 0} = - 1$

From equation (ii) we get, $y$ (In $\sqrt{3} ) = - \dfrac{1}{\sqrt{3}}$

So, 6\left[y ' \left(0\right) + ( y\right) In \left\sqrt{3}{)}^{2}\right] = 6\left[- 1 + \dfrac{1}{3}\right] = - 4.