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(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x^2-1}$ and $Q

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$19-6 \log_{e} 3$

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(D) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x^2-1}$ and $Q(x) = \left(\frac{x-1}{x+1}\right)^{1/2}$.The Integrating Factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int \frac{1}{x^2-1} dx} = e^{\frac{1}{2} \log \left| \frac{x-1}{x+1} \right|} = \left( \frac{x-1}{x+1} \right)^{1/2}$.The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$.$y \left( \frac{x-1}{x+1} \right)^{1/2} = \int \left( \frac{x-1}{x+1} \right)^{1/2} \cdot \left( \frac{x-1}{x+1} \right)^{1/2} dx = \int \frac{x-1}{x+1} dx$.$y \left( \frac{x-1}{x+1} \right)^{1/2} = \int \left( 1 - \frac{2}{x+1} \right) dx = x - 2 \log_{e} |x+1| + C$.Since the curve passes through $\left(2, \frac{1}{\sqrt{3}}\right)$,we substitute $x=2$ and $y=\frac{1}{\sqrt{3}}$:$\frac{1}{\sqrt{3}} \left( \frac{2-1}{2+1} \right)^{1/2} = 2 - 2 \log_{e} 3 + C \implies \frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}} = 2 - 2 \log_{e} 3 + C \implies \frac{1}{3} = 2 - 2 \log_{e} 3 + C$.$C = \frac{1}{3} - 2 + 2 \log_{e} 3 = 2 \log_{e} 3 - \frac{5}{3}$.Now,for $x=8$:$y(8) \left( \frac{8-1}{8+1} \right)^{1/2} = 8 - 2 \log_{e} 9 + 2 \log_{e} 3 - \frac{5}{3}$.$y(8) \left( \frac{7}{9} \right)^{1/2} = 8 - 4 \log_{e} 3 + 2 \log_{e} 3 - \frac{5}{3} = \frac{19}{3} - 2 \log_{e} 3$.$y(8) \cdot \frac{\sqrt{7}}{3} = \frac{19 - 6 \log_{e} 3}{3}$.$\sqrt{7} y(8) = 19 - 6 \log_{e} 3$.

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

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