Answer
The solutions are ${\bf{r}}\left( t \right) = {{\rm{e}}^{2t}}{\bf{c}}$,
where ${\bf{c}} = \left( {{c_1},{c_2},{c_3}} \right)$, a constant vector.
Work Step by Step
We have ${\bf{r}}'\left( t \right) = 2{\bf{r}}\left( t \right)$, where ${\bf{r}}\left( t \right)$ is a vector-valued function in 3-space. Let the components of ${\bf{r}}\left( t \right)$ be ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right)$.
So, ${\bf{r}}'\left( t \right) = \left( {x'\left( t \right),y'\left( t \right),z'\left( t \right)} \right)$.
Thus, ${\bf{r}}'\left( t \right) = 2{\bf{r}}\left( t \right)$ becomes
$\left( {x'\left( t \right),y'\left( t \right),z'\left( t \right)} \right) = \left( {2x\left( t \right),2y\left( t \right),2z\left( t \right)} \right)$
We obtain a system of equations:
$x'\left( t \right) = 2x\left( t \right)$, ${\ \ }$ $y'\left( t \right) = 2y\left( t \right)$, ${\ \ }$ $z'\left( t \right) = 2z\left( t \right)$
Evaluate $x'\left( t \right) = 2x\left( t \right)$.
Write $x'\left( t \right) = \frac{{dx\left( t \right)}}{{dt}} = 2x\left( t \right)$. So,
$\smallint \frac{{dx\left( t \right)}}{{x\left( t \right)}} = 2\smallint {\rm{d}}t$,
$\ln x\left( t \right) = 2t$,
$x\left( t \right) = {c_1}{{\rm{e}}^{2t}}$,
where ${c_1}$ is constant of integration.
By symmetry we also obtain $y\left( t \right) = {c_2}{{\rm{e}}^{2t}}$ and $z\left( t \right) = {c_3}{{\rm{e}}^{2t}}$. Hence,
${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right),z\left( t \right)} \right) = \left( {{c_1}{{\rm{e}}^{2t}},{c_2}{{\rm{e}}^{2t}},{c_3}{{\rm{e}}^{2t}}} \right)$
${\bf{r}}\left( t \right) = {{\rm{e}}^{2t}}\left( {{c_1},{c_2},{c_3}} \right)$
Write ${\bf{c}} = \left( {{c_1},{c_2},{c_3}} \right)$.
So, the solutions are ${\bf{r}}\left( t \right) = {{\rm{e}}^{2t}}{\bf{c}}$.
