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