Answer
The general solution of the position vector:
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t} - t,3t - \sin t,2t - \cos t} \right) + {{\bf{c}}_2}$,
where ${{\bf{c}}_2}$ is a constant vector.
The solution with the given initial condition:
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t} - t,3t - \sin t,2t - \cos t + 2} \right)$
Work Step by Step
We have the differential equation:
${\bf{r}}{\rm{''}}\left( t \right) = \left( {{{\rm{e}}^t},\sin t,\cos t} \right)$
We find the general solution of the velocity vector by integration:
${\bf{r}}'\left( t \right) = \mathop \smallint \limits_{}^{} \left( {{{\rm{e}}^t},\sin t,\cos t} \right){\rm{d}}t = \left( {{{\rm{e}}^t}, - \cos t,\sin t} \right) + {{\bf{c}}_1}$
Using the initial condition ${\bf{r}}'\left( 0 \right) = \left( {0,2,2} \right)$, we obtain
${\bf{r}}'\left( 0 \right) = \left( {1, - 1,0} \right) + {{\bf{c}}_1} = \left( {0,2,2} \right)$
Therefore, ${{\bf{c}}_1} = \left( { - 1,3,2} \right)$. So, the solution with the given initial condition is
${\bf{r}}'\left( t \right) = \left( {{{\rm{e}}^t}, - \cos t,\sin t} \right) + \left( { - 1,3,2} \right)$
${\bf{r}}'\left( t \right) = \left( {{{\rm{e}}^t} - 1,3 - \cos t,2 + \sin t} \right)$
We find the general solution of the position vector by integration:
${\bf{r}}\left( t \right) = \mathop \smallint \limits_{}^{} \left( {{{\rm{e}}^t} - 1,3 - \cos t,2 + \sin t} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t} - t,3t - \sin t,2t - \cos t} \right) + {{\bf{c}}_2}$
Using the initial condition ${\bf{r}}\left( 0 \right) = \left( {1,0,1} \right)$, we obtain
${\bf{r}}\left( 0 \right) = \left( {1,0, - 1} \right) + {{\bf{c}}_2} = \left( {1,0,1} \right)$
Therefore, ${{\bf{c}}_2} = \left( {0,0,2} \right)$. So, the solution with the given initial condition is
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t} - t,3t - \sin t,2t - \cos t} \right) + \left( {0,0,2} \right)$
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t} - t,3t - \sin t,2t - \cos t + 2} \right)$
