Official solution
\[{{dy} \over {dx}} + {1 \over {{x^2} - 1}}y = \sqrt {{{x - 1} \over {x + 1}}} ,\,x > 1\]
Integrating factor I.F. \[ = {e^{\int {{1 \over {{x^2} - 1}}dx} }} = {e^{{1 \over 2}\ln \left| {{{x - 1} \over {x + 1}}} \right|}}\]
\[ = \sqrt {{{x - 1} \over {x + 1}}} \]
Solution of differential equation
\[y\sqrt {{{x - 1} \over {x + 1}}} = \int {{{x - 1} \over {x + 1}}dx = \int {\left( {1 - {2 \over {x + 1}}} \right)dx} } \]
\[y\sqrt {{{x - 1} \over {x + 1}}} = x - 2\ln |x + 1| + C\]
Curve passes through \[\left( {2,\sqrt {{1 \over 3}} } \right)\]
\[{1 \over {\sqrt 3 }} \times {1 \over {\sqrt 3 }} = 2 - 2\ln 3 + C\]
\[C = 2\ln 3 - {5 \over 3}\]
\[y(8) \times {{\sqrt 7 } \over 3} = 8 - 2\ln 9 + 2\ln 3 - {5 \over 3}\]
\[\sqrt 7 \,.\,y(8) = 19 - 6\ln 3\]