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Step-by-step Solution
Step 1: Write down the differential equation
The given differential equation is
$$
\frac{dy}{dx} \;+\; \frac{1}{x^2 - 1} \, y \;=\; \sqrt{\frac{x - 1}{x + 1}}, \quad x > 1.
$$
Step 2: Identify the integrating factor (I.F.)
The integrating factor I.F. is given by
$$
e^{\displaystyle \int \frac{1}{x^2 - 1}\,dx}.
$$
We compute:
$$
\int \frac{1}{x^2 - 1}\,dx
= \int \frac{1}{(x-1)(x+1)}\,dx
= \frac{1}{2}\,\ln\left|\frac{x-1}{x+1}\right|.
$$
Therefore, the integrating factor is
$$
\text{I.F.} \;=\; e^{\frac{1}{2}\ln\left|\frac{x - 1}{x + 1}\right|}
\;=\; \sqrt{\frac{x - 1}{x + 1}}.
$$
Step 3: Multiply the differential equation by the I.F.
Multiplying both sides of the original differential equation by
$$
\sqrt{\frac{x - 1}{x + 1}},
$$
we obtain
$$
\sqrt{\frac{x - 1}{x + 1}}\;\frac{dy}{dx} \;+\; \frac{1}{x^2 - 1}\,\sqrt{\frac{x - 1}{x + 1}}\,y
= \sqrt{\frac{x - 1}{x + 1}}\;\sqrt{\frac{x - 1}{x + 1}}.
$$
The left side simplifies to the derivative of
$$
y \,\sqrt{\frac{x - 1}{x + 1}}.
$$
Hence, we can write
$$
\frac{d}{dx}\Bigl(y \,\sqrt{\frac{x - 1}{x + 1}}\Bigr)
= \frac{x - 1}{x + 1}.
$$
Step 4: Integrate both sides
We now integrate the right-hand side:
$$
\int \frac{x - 1}{x + 1}\,dx.
$$
Rewrite
$$
\frac{x - 1}{x + 1} \;=\; 1 - \frac{2}{x + 1}.
$$
So,
$$
\int \biggl(1 - \frac{2}{x + 1}\biggr)\,dx
= \int 1\,dx \;-\; 2 \int \frac{1}{x + 1}\,dx
= x \;-\; 2\,\ln|x + 1| \;+\; C.
$$
Thus,
$$
y \,\sqrt{\frac{x - 1}{x + 1}}
= x \;-\; 2\,\ln|x + 1|\;+\; C.
$$
Step 5: Apply the initial condition
The solution curve passes through the point $(2,\sqrt{1/3})$. Hence, when $x=2$, $y=\sqrt{\frac{1}{3}}$. We substitute these values into
$$
y \,\sqrt{\frac{x - 1}{x + 1}}
= x - 2\,\ln|x + 1| + C.
$$
For $x=2$:
$$
\sqrt{\frac{1}{3}} \,\sqrt{\frac{2 - 1}{2 + 1}}
= 2 \;-\; 2\,\ln(3) \;+\; C.
$$
Notice that
$$
\sqrt{\frac{1}{3}} \,\sqrt{\frac{1}{3}}
= \frac{1}{3}.
$$
So the left side becomes $\tfrac{1}{3}$. Therefore,
$$
\frac{1}{3}
= 2 - 2\,\ln(3) + C
\quad\Rightarrow\quad
C = 2\,\ln(3) - \frac{5}{3}.
$$
Step 6: Write the general solution including the constant
Thus, the particular solution is
$$
y \,\sqrt{\frac{x - 1}{x + 1}}
= x \;-\; 2\,\ln|x + 1| \;+\; 2\,\ln(3) \;-\; \frac{5}{3}.
$$
Step 7: Find $y(8)$
We now substitute $x=8$:
$$
y(8)\,\sqrt{\frac{8 - 1}{8 + 1}}
= 8 \;-\; 2\,\ln(9) \;+\; 2\,\ln(3) \;-\; \frac{5}{3}.
$$
Observe that $\sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$, and $\ln(9)=2\ln(3)$. Thus,
$$
y(8)\,\frac{\sqrt{7}}{3}
= 8 \;-\; 2\,\bigl(2\,\ln(3)\bigr) \;+\; 2\,\ln(3) \;-\; \frac{5}{3}.
$$
Simplify the right side:
$$
8 \;-\; 4\,\ln(3) \;+\; 2\,\ln(3) \;-\; \frac{5}{3}
= 8 \;-\; 2\,\ln(3) \;-\; \frac{5}{3}.
$$
Combine the constants:
$$
8 - \frac{5}{3} = \frac{24 - 5}{3} = \frac{19}{3}.
$$
So the expression becomes
$$
\frac{19}{3} - 2\,\ln(3).
$$
Therefore,
$$
y(8)\,\frac{\sqrt{7}}{3}
= \frac{19}{3} - 2\,\ln(3).
$$
Multiply both sides by $\frac{3}{\sqrt{7}}$ to get $y(8)\,\sqrt{7}$:
$$
y(8)\,\sqrt{7}
= \biggl(\frac{19}{3} - 2\,\ln(3)\biggr)\times\frac{3}{\sqrt{7}}
= \frac{19 - 6\,\ln(3)}{\sqrt{7}}\times \sqrt{7}
= 19 - 6\,\ln(3).
$$
Final Answer
$$
\sqrt{7}\,y(8) \;=\; 19 - 6\,\ln(3).
$$
Hence, the correct answer is
$\;19 - 6 \log_{e}(3)\;$.
