Let y = y x be the solution curve of the differential equation d y d x + 1 x 2 - 1 y = x - 1 x + 1 1 2 , x…

Given,

$\frac{dy}{dx}+\frac{y}{x^2-1}={\left(\frac{x-1}{x+1}\right)}^{\frac{1}{2}}$ is a linear differential equation.

Here $\mathrm{I}.\mathrm{F}.=e^{\int \frac{dx}{x^2-1}}=e^{\frac{1}{2}\ln \left(\frac{x-1}{x+1}\right)}=\sqrt{\frac{x-1}{x+1}}$

General solution will be $y\sqrt{\frac{x-1}{x+1}}=\int \frac{x-1}{x+1}dx+C$

$\Rightarrow y\sqrt{\frac{x-1}{x+1}}=x-2\ln \left(x+1\right)+c$

Given $y\left(2\right)=\frac{1}{\sqrt{3}}$

$\Rightarrow c=2\ln 3-\frac{5}{3}$

So, the equation of the curve is  $y\sqrt{\frac{x-1}{x+1}}=x-2\ln \left(x+1\right)+2\ln 3-\frac{5}{3}$

Now putting $x=8$, we get

$\frac{\sqrt{7}y\left(8\right)}{3}=8-4\ln 3+2\ln 3-\frac{5}{3}$

$\Rightarrow \sqrt{7}y\left(8\right)=19-6\ln 3$