I have a question, the first part of which I think I have solved, but the solution seems suspicious, and the second part which I have not. Please would you help?
The first part of the question is "What are the orbits and stabilisers of the group action $\Bbb R\ \times \Bbb R^2 \to \Bbb R^2$; $(t, (x,y)) \to (e^tx, e^{-t}y)$?"
I believe the orbit of $(x,y)$ is just $(e^tx, e^{-t}y)$ and although I note that this ordered pair has the same product as the ordered pair in the domain, otherwise I'm not sure what point the question is trying to make - am I missing something?
I think the stabiliser of $(x,y)$ is {$0$} iff $(x, y)$ is not $(0,0)$ and $\Bbb R$ in that particular case. Is this correct?
Finally, the last part of the question is "What is the differential equation satisfied by each of the orbits?". I have not properly attempted this because I was stuck on the first part. If possible, could I have hints to this rather than a solution?
Thank you.
In your solution for what the orbit might be, you still have $t$ as a parameter: that can't be right.
The orbit is, by definition $\{(e^t x, e^{-t}y), t\in \mathbb{R}\}$. Now clearly, you are meant to give a more explicit description of the orbit.
To give this more explicit description, you can notice a few things: as you noticed, if $(x',y')$ is in the orbit, $x'y' = xy$.
Moreover, defining the sign of $x$ to be $1$ if $x>0$, $0$ if $x=0$, and $-1$ if $x<0$, one has that $x$ and $x'$ have the same sign, and so do $y, y'$.
Conversely, assume these two conditions are satisfied.
If $x\neq 0$, consider $\frac{x'}{x}$. As $x'$ and $x$ have the same sign, this is $>0$, and so is of the form $e^t$ for some $t\in \mathbb{R}$. Moreover, $x' \neq 0$ and $x'y' = xy$ so $y' = \frac{x}{x'} y = e^{-t}y$ and so $(x',y')$ is in the orbit of $(x,y)$.
If $x=0, y= 0$, clearly the orbit of $(x,y)$ is $\{(0,0)\}$.
If $x=0, y \neq 0$ then do the same as before but with $y$ instead of $x$, and you find again that $(x',y')$ is in the orbit of $(x,y)$.
So the orbit of $(x,y)$ is $\{(x',y') \mid sgn(x)=sgn(x'), sgn(y)=sgn(y'), xy=x'y'\}$
As for the sabilisers, it depends on whether the two are $0$ or not. If the two are $0$, as you note, the stabiliser is $\mathbb{R}$.
Similarly, if for instance $x\neq 0$, $e^t x = x$ has only one solution in $t$ ($t=0$), so that the stabiliser of $x$ is $\{0\}$, as you noticed.
The second question asks for a differential equation, which makes sense because by definition the orbit is parametrized by $t$. Notice that $\frac{\mathrm{d}}{\mathrm{d}t} (e^t x, e^{-t}y) = (e^t x, - e^{-t}y)$ and so $\frac{\mathrm{d}^2}{\mathrm{d}t^2} (e^t x, e^{-t}y) = (e^t x, e^{-t}y)$, so denoting $\gamma(t)= (e^t x, e^{-t}y)$, one has $\frac{\mathrm{d}^2}{\mathrm{d}t^2} \gamma (t) = \gamma(t)$. I assume that this is what is meant by "the differential equation of the orbit".