Answer
$$y\left( t \right) = {t^3} - 2{t^2} + 10t + 20$$
Work Step by Step
$$\eqalign{
& y'\left( t \right) = 3{t^2} - 4t + 10 \cr
& {\text{integrate both sides }} \cr
& \int {y'\left( t \right)} = \int {\left( {3{t^2} - 4t + 10} \right)dt} \cr
& y\left( t \right) = {t^3} - 2{t^2} + 10t + C \cr
& {\text{the initial condition }}y\left( 0 \right) = 20{\text{ implies that}} \cr
& 20 = {\left( 0 \right)^3} - 2{\left( 0 \right)^2} + 10t + C \cr
& C = 20 \cr
& {\text{so the solution of the initial value problem is}} \cr
& y\left( t \right) = {t^3} - 2{t^2} + 10t + 20 \cr} $$
