Answer
$$ y = \sqrt {{x^2} - 4} - 2{\sec ^{ - 1}}\left( {\frac{x}{2}} \right)$$
Work Step by Step
$$\eqalign{
& x\frac{{dy}}{{dx}} = \sqrt {{x^2} - 4},\,\,\,\,\,\,\,x \geqslant 2,\,\,\,\,\,\,y\left( 2 \right) = 0 \cr
& {\text{Separate variables}} \cr
& dy = \frac{{\sqrt {{x^2} - 4} }}{x}dx \cr
& {\text{then}} \cr
& y = \int {\frac{{\sqrt {{x^2} - 4} }}{x}} dx \cr
& {\text{We set}}{\text{, }}x = 2\sec \theta,{\text{ }}dx = 2\sec \theta \tan \theta d\theta \cr
& {\text{With these substitutions}}{\text{, we have}} \cr
& y = \int {\frac{{\sqrt {{{\left( {2\sec \theta } \right)}^2} - 4} }}{{2\sec \theta }}} \left( {2\sec \theta \tan \theta d\theta } \right) \cr
& y = \int {\frac{{\sqrt {4{{\sec }^2}\theta - 4} }}{{2\sec \theta }}} \left( {2\sec \theta \tan \theta d\theta } \right) \cr
& y = \int {\sqrt {4\left( {{{\sec }^2}\theta - 1} \right)} } \tan \theta d\theta \cr
& {\text{use the fundamental identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& y = \int {2\sqrt {{{\tan }^2}\theta } } \tan \theta d\theta \cr
& {\text{simplifying}} \cr
& y = 2\int {{{\tan }^2}\theta } d\theta \cr
& y = 2\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{integrate}} \cr
& y = 2\tan \theta - 2\theta + C \cr
& {\text{write in terms of }}x, \cr
& y = \sqrt {{x^2} - 4} - 2{\sec ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr
& {\text{use initial condition }}y\left( 2 \right) = 0 \cr
& 0 = \sqrt {{{\left( 2 \right)}^2} - 4} - 2{\sec ^{ - 1}}\left( {\frac{2}{2}} \right) + C \cr
& 0 = \sqrt 0 - 2\left( 0 \right) + C \cr
& C = 0 \cr
& {\text{then}} \cr
& y = \sqrt {{x^2} - 4} - 2{\sec ^{ - 1}}\left( {\frac{x}{2}} \right) \cr} $$
