Answer
$${\bf{y}}\left( t \right) = \left( {{t^2} + 1} \right){\bf{i}} + \left( {{t^3} - 1} \right){\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{y}}'\left( t \right) = 2t{\bf{i}} + 3{t^2}{\bf{j}},{\text{ }}{\bf{y}}\left( 0 \right) = {\bf{i}} - {\bf{j}} \cr
& {\text{Calculate }}{\bf{y}}\left( t \right) \cr
& {\bf{y}}\left( t \right) = \int {{\bf{y}}'\left( t \right)} dt \cr
& {\bf{y}}\left( t \right) = \int {\left( {2t{\bf{i}} + 3{t^2}{\bf{j}}} \right)} dt \cr
& {\bf{y}}\left( t \right) = {t^2}{\bf{i}} + {t^3}{\bf{j}} + {\bf{C}} \cr
& {\text{Use the initial condition }}{\bf{y}}\left( 0 \right) = {\bf{i}} - {\bf{j}} \cr
& {\bf{y}}\left( 0 \right) = {\left( 0 \right)^2}{\bf{i}} + {\left( 0 \right)^3}{\bf{j}} + {\bf{C}} \cr
& {\bf{i}} - {\bf{j}} = {\bf{C}} \cr
& {\text{Therefore,}} \cr
& {\bf{y}}\left( t \right) = {t^2}{\bf{i}} + {t^3}{\bf{j}} + {\bf{i}} - {\bf{j}} \cr
& {\bf{y}}\left( t \right) = {t^2}{\bf{i}} + {\bf{i}} + {t^3}{\bf{j}} - {\bf{j}} \cr
& {\bf{y}}\left( t \right) = \left( {{t^2} + 1} \right){\bf{i}} + \left( {{t^3} - 1} \right){\bf{j}} \cr} $$
