Answer
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^{ - t}},2{{\rm{e}}^{ - t}}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$. So, ${\bf{r}}'\left( t \right) = \left( {x'\left( t \right),y'\left( t \right)} \right)$.
The requirement is ${\bf{r}}'\left( t \right) = - {\bf{r}}\left( t \right)$. So,
$\left( {x'\left( t \right),y'\left( t \right)} \right) = - \left( {x\left( t \right),y\left( t \right)} \right)$
In component forms, we get
$x'\left( t \right) = - x\left( t \right)$ ${\ \ }$ and ${\ \ }$ $y'\left( t \right) = - y\left( t \right)$
Ignoring the $t$ parameter for convenience, we have
$\frac{{dx}}{{dt}} = - x$ ${\ \ }$ and ${\ \ }$ $\frac{{dy}}{{dt}} = - y$
$\frac{{dx}}{x} = - dt$ ${\ \ }$ and ${\ \ }$ $\frac{{dy}}{y} = - dt$
Integrating gives
$\smallint \frac{1}{x}{\rm{d}}x = - \smallint {\rm{d}}t$ ${\ \ }$ and ${\ \ }$ $\smallint \frac{1}{y}{\rm{d}}y = - \smallint {\rm{d}}t$
$\ln x = - t + {x_0}$ ${\ \ }$ and ${\ \ }$ $\ln y = - t + {y_0}$
At $t=0$, we have ${\bf{r}}\left( 0 \right) = \left( {1,2} \right)$. So,
${x_0} = \ln 1 = 0$ ${\ \ }$ and ${\ \ }$ ${y_0} = \ln 2$
Thus,
$\ln x = - t$ ${\ \ }$ and ${\ \ }$ $\ln y = - t + \ln 2$
$x = {{\rm{e}}^{ - t}}$ ${\ \ }$ and ${\ \ }$ $y = {{\rm{e}}^{ - t + \ln 2}} = 2{{\rm{e}}^{ - t}}$
Hence, ${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^{ - t}},2{{\rm{e}}^{ - t}}} \right)$.
