Answer
The solutions are ${\bf{r}}\left( t \right) = \left( {t - 1} \right){\bf{v}} + {\bf{w}}$
Work Step by Step
We have ${\bf{r}}'\left( t \right) = {\bf{v}}$ with initial condition ${\bf{r}}\left( 1 \right) = {\bf{w}}$, where ${\bf{v}}$ and ${\bf{w}}$ are constant vectors in ${\mathbb{R}^3}$.
The solution is obtained by integrating ${\bf{r}}'\left( t \right)$:
${\bf{r}}\left( t \right) = \smallint {\bf{v}}{\rm{d}}t = {\bf{v}}t + {\bf{c}}$,
where ${\bf{c}}$ is a constant vector.
Since ${\bf{r}}\left( 1 \right) = {\bf{w}}$, we have ${\bf{r}}\left( 1 \right) = {\bf{v}} + {\bf{c}} = {\bf{w}}$. So, ${\bf{c}} = {\bf{w}} - {\bf{v}}$.
${\bf{r}}\left( t \right) = {\bf{v}}t + {\bf{c}} = {\bf{v}}t + {\bf{w}} - {\bf{v}}$
Hence, the solutions are
${\bf{r}}\left( t \right) = \left( {t - 1} \right){\bf{v}} + {\bf{w}}$
