Answer
$$y = \sqrt {t + \ln t + 15} $$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = \frac{{t + 1}}{{2ty}},\,\,\,\,\,\,y\left( 1 \right) = 4 \cr
& {\text{Rewrite}} \cr
& \frac{{dy}}{{dt}} = \frac{{t + 1}}{{2ty}} \cr
& {\text{Separate the variables}} \cr
& 2ydy = \frac{{t + 1}}{t}dt \cr
& {\text{Integrating}} \cr
& \int {2y} dy = \int {\left( {1 + \frac{1}{t}} \right)} dt \cr
& {y^2} = t + \ln \left| t \right| + C \cr
& {\text{Use the initial condition }}\,y\left( 1 \right) = 4 \cr
& {\left( 4 \right)^2} = 1 + \ln \left| 1 \right| + C \cr
& C = 15 \cr
& ,{\text{ then}} \cr
& {y^2} = t + \ln \left| t \right| + 15 \cr
& {\text{Solve for }}y \cr
& y = \sqrt {t + \ln t + 15} \cr} $$
