Answer
$$ y = {\sec ^{ - 1}}x + \frac{2}{3}\pi $$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{1}{{x\sqrt {{x^2} - 1} }},\,\,\,\,x > 1;\,\,\,\,\,\,y\left( 2 \right) = \pi \cr
& {\text{separating the variables, we get:}} \cr
& dy = \left( {\frac{1}{{x\sqrt {{x^2} - 1} }}} \right)dx \cr
& {\text{integrate both sides}} \cr
& y = \int {\left( {\frac{1}{{x\sqrt {{x^2} - 1} }}} \right)} dx \cr
& y = {\sec ^{ - 1}}x + C\,\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{use initial condition }}y\left( 2 \right) = \pi \cr
& \pi = {\sec ^{ - 1}}\left( 2 \right) + C \cr
& \pi = \frac{\pi }{3} + C \cr
& C = \pi - \frac{\pi }{3} = \frac{2}{3}\pi \cr
& \cr
& {\text{then}}{\text{, substitute }}C = \frac{2}{3}\pi {\text{ in }}\left( {\bf{1}} \right) \cr
& y = {\sec ^{ - 1}}x + \frac{2}{3}\pi \cr} $$
