Answer
$$y\left( t \right) = {t^2} + 4\ln \left| t \right| + 1$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = \left( {2{t^2} + 4} \right)/t \cr
& y'\left( t \right) = \frac{{2{t^2}}}{t} + \frac{4}{t} \cr
& {\text{integrate both sides }} \cr
& \int {y'\left( t \right)} = \int {\left( {2t + \frac{4}{t}} \right)dt} \cr
& y\left( t \right) = {t^2} + 4\ln \left| t \right| + C \cr
& {\text{the initial condition }}y\left( 1 \right) = 2{\text{ implies that}} \cr
& 2 = {\left( 1 \right)^2} + 4\ln \left| 1 \right| + C \cr
& C = 1 \cr
& {\text{so the solution of the initial value problem is}} \cr
& y\left( t \right) = {t^2} + 4\ln \left| t \right| + 1 \cr} $$
