Answer
$$y = 2{e^{2\arcsin \left( {\frac{x}{4}} \right)}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{2y}}{{\sqrt {16 - {x^2}} }} \cr
& {\text{Separate the variables}} \cr
& \frac{1}{y}dy = \frac{2}{{\sqrt {16 - {x^2}} }}dx \cr
& {\text{Integrate both sides}} \cr
& \int {\frac{1}{y}dy} = 2\int {\frac{1}{{\sqrt {16 - {x^2}} }}} dx \cr
& {\text{By Theorem 5}}{\text{.17 }}\int {\frac{{du}}{{\sqrt {{a^2} - {u^2}} }} = \arcsin \left( {\frac{u}{a}} \right) + C} \cr
& \ln \left| y \right| = 2\arcsin \left( {\frac{x}{4}} \right) + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}y\left( 0 \right) = 2 \cr
& \ln \left| 2 \right| = 2\arcsin \left( {\frac{0}{4}} \right) + C \cr
& C = \ln 2 \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& \ln \left| y \right| = 2\arcsin \left( {\frac{x}{4}} \right) + \ln 2 \cr
& {\text{Solve for }}y \cr
& y = 2{e^{2\arcsin \left( {\frac{x}{4}} \right)}} \cr} $$
