Let $$y = y(x)$$ be the solution curve of the differential equation $$\frac{dy}{dx} + \frac{1}{x^2-1}y = \left(\frac{x-1}{x+1}\right)^{1/2}$$, $$x > 1$$ passing through the point $$\left(2, \sqrt{\frac{1}{3}}\right)$$. Then $$\sqrt{7}y(8)$$ is equal to

We have the differential equation $$\frac{dy}{dx} + \frac{1}{x^2-1}\,y = \left(\frac{x-1}{x+1}\right)^{1/2}$$, with $$x > 1$$, passing through $$\left(2, \sqrt{\frac{1}{3}}\right)$$.

This is a first-order linear ODE. The integrating factor is $$\mu = e^{\int \frac{1}{x^2-1}\,dx}$$. We decompose $$\frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} = \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right)$$.

So $$\int \frac{dx}{x^2-1} = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|$$, and the integrating factor is:

$$\mu = e^{\frac{1}{2}\ln\frac{x-1}{x+1}} = \left(\frac{x-1}{x+1}\right)^{1/2}$$

(since $$x > 1$$, both $$x-1 > 0$$ and $$x+1 > 0$$).

Multiplying both sides by $$\mu$$:

$$\frac{d}{dx}\left[y\left(\frac{x-1}{x+1}\right)^{1/2}\right] = \left(\frac{x-1}{x+1}\right)^{1/2} \cdot \left(\frac{x-1}{x+1}\right)^{1/2} = \frac{x-1}{x+1}$$

$$= 1 - \frac{2}{x+1}$$

Integrating both sides:

$$y\sqrt{\frac{x-1}{x+1}} = x - 2\ln(x+1) + C$$

Now we apply the initial condition $$\left(2, \sqrt{\frac{1}{3}}\right)$$:

$$\sqrt{\frac{1}{3}} \cdot \sqrt{\frac{1}{3}} = 2 - 2\ln 3 + C$$

$$\frac{1}{3} = 2 - 2\ln 3 + C$$

$$C = \frac{1}{3} - 2 + 2\ln 3 = -\frac{5}{3} + 2\ln 3$$

So $$y\sqrt{\frac{x-1}{x+1}} = x - 2\ln(x+1) - \frac{5}{3} + 2\ln 3$$.

At $$x = 8$$:

$$y(8)\sqrt{\frac{7}{9}} = 8 - 2\ln 9 - \frac{5}{3} + 2\ln 3$$

$$y(8) \cdot \frac{\sqrt{7}}{3} = 8 - 2\ln 9 + 2\ln 3 - \frac{5}{3}$$

Now $$2\ln 3 - 2\ln 9 = 2\ln 3 - 4\ln 3 = -2\ln 3$$. So:

$$y(8) \cdot \frac{\sqrt{7}}{3} = 8 - 2\ln 3 - \frac{5}{3} = \frac{24 - 5}{3} - 2\ln 3 = \frac{19}{3} - 2\ln 3$$

$$\sqrt{7}\,y(8) = 3\left(\frac{19}{3} - 2\ln 3\right) = 19 - 6\ln 3 = 19 - 6\log_e 3$$

Hence, the correct answer is Option D.