If y = y x is the solution of the differential equation 1 + e 2 x d y d x + 2 1 + y 2 e x = 0 and y 0 = 0 ,…

Given,

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

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

Now integrating both side we get,

$\Rightarrow \int \frac{dy}{1+y^2}+\int \frac{2e^x}{1+e^{2x}}dx=\int 0$

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

$\because y\left(0\right)=0$

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

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

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

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