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