Chapter 12 - Vector-Valued Functions - 12.2 Calculus Of Vector-Valued Functions - Exercises Set 12.2 - Page 857: 47

$${\bf{y}}\left( t \right) = \left( {\frac{{{t^2}}}{2} + 2} \right){\bf{i}} + \left( {{e^t} - 1} \right){\bf{j}}$$

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